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THE  ELEMENTS 

OF  ^ 

Plane  and  Solid  Geometey. 

WITH  NUMEEOUS  EXERCISES. 


BY 


EDWARD  A.  BOWSER,  LL.D, 

Professor  of  Mathematics  and  Ekginekring  in  Rutgers  College. 


( 


vT' 


NEW  YORK: 

D.  VAN  NOSTRAND   COMPANY,  Publishers, 

23  MuRiiAY  AND  27  WARiiEif  Stueets, 

1890. 


CopyRiOHT,  1890, 
By  E.  a.  Bowsk^. 


PREFACE. 


"TN  the  present  work  on  Elementary  Geometry,  or  what  is  com- 
-*-  monly  known  as  the  Euclidian  Geometry,  it  is  aimed  to  com- 
bine the  excellencies  of  Euclid  with  those  of  the  best  modern 
writers,  especially  of  Legendre,  and  Rouche  and  Comberousse. 
Many  of  the  demonstrations  are  those  of  Euclid,  with  minor  changes 
frequently  introduced,  and  the  syllogistic  form  is  retained  through- 
out ;  but  the  arrangement  is  quite  different. 

Many  objections  have  been  made  against  Euclid.  His  definitions 
are  not  all  of  them  the  best,  nor  are  they  in  their  proper  places. 
His  treatment  of  angles  is  deficient.  His  arrangement  of  the  propo- 
sitions is  often  poor,  mixing  straight  lines,  angles,  triangles,  etc., 
without  any  regular  classification ;  and  his  demonstrations  are 
sometimes  cumbersome  and  prolix. 

Nevertheless,  although  numerous  attempts  have  been  made  to 
improve  upon  Euclid,  it  still  remains  the  great  model,  the  unrivalled 
original,  on  which  is  founded  the  whole  system  of  elementary 
Geometry.  Perhaps  a  more  finished  specimen  of  exact  logic  has 
never  been  produced  than  that  of  the  old  Greek  Geometer. 

In  the  present  treatise  it  is  desired  to  effect  two  objects:  (1)  to 
teach  geometric  truths  ;  (2)  to  discipline  and  invigorate  the  mind, 
to  train  it  to  habits  of  clear  and  consecutive  reasoning.  Accord- 
ingly, more  numerous  propositions  have  been  given,  and  the 
demonstrations  made  more  complete,  than  either  object  alone  would 
seem  to  demand. 

In  each  proposition  is  a  distinct  statement,  of  what  is  given,  of 
what  is  required,  and  of  the  proof.  Each  assertion  in  the  proof 
begins  a  new  line,  and  is  accompanied  by  a  reference  to  the  preced- 
ing principle  on  which  the  assertion  depends.  These  references  are 
quoted  two  or'^three  times  in  small  type,  and  afterwards  referred  to 
only  by  number.  The  student  should  alicays  be  ready,  if  required,  to 
quote  the  proper  reference,  and  to  show  its  application.  The  text  is  so 
arranged  that  the  enunciation,  figure,  and  proof  of  each  proposition 

183623  "' 


IV  PREFACE. 

are  in  view  together ;  and  notes  are  directly  appended  to  the  propo- 
sitions to  which  they  refer. 

A  few  symbols  and  abbreviations  of  words  have  been  freely  used, 
but  only  such  as  have  long  been  employed  by  mathematicians,  and 
are  recognized  by  the  majority  of  teachers. 

In  the  figures  the  given  lines  are  represented  by  full  lines,  those 
which  are  added  to  aid  in  the  demonstration,  by  sfwrt-dotted  lines, 
and  the  remlting  lines  by  dashed,  or  long-dotted  lines.  In  solid 
geometry  the  dotted  lines  represent  those  which  a  solid  body  would 
conceal. 

The  propositions  marked  with  a  star  may  be  omitted  without 
interfering  with  the  continuity  of  the  work,  but  the  omission  is  not 
recommended. 

The  exercises  distributed  through  the  text  are  quite  easy,  and 
may  all  be  worked  out  by  the  average  student ;  those  at  the  end  of 
each  book  are  more  advanced,  and  have  been  carefully  graded,  with 
hints  appended  to  many  of  the  more  diflScult  ones. 

It  is  only  in  original  demonstration  that  the  student  can  acquire 
mental  power.  More  discipline  is  gained  in  working  out  one 
demonstration,  without  aid,  than  by  learning  a  number  of  them 
that  are  given  by  others.  A  student  can  never  really  compreliend  a 
subject  if  he  only  tries  to  understand  and  remember  what  the  book 
says.  The  subject  can  become  known  to  him  only  by  his  thinking 
upon  it.  To  develop  the  power  of  independent  thought  in  the 
student,  is  the  most  important  part  of  the  teacher's  work,  and  it  is 
the  most  difficult. 

In  preparing  this  work  I  have  consulted  some  of  the  best  Ameri- 
can and  English  books  on  Euclidian  Geometry,  and  am  especially 
indebted  to  the  text-book  recently  published  bj'  the  English  Asso- 
ciation for  the  Improvement  of  Geometrical  Teaching.  I  have  also 
derived  assistance  from  a  number  of  French  works,  especially  from 
those  of  Catalan,  Briot,  and  Legendre.  while  the  Traite  de  Geomelrie 
by  Rouche  and  De  Comberousse  has  been  my  constant  companion. 

It  remains  for  me  to  express  my  thanks  to  Prof.  R.  W.  Prentiss, 
of  the  Nautical  Almanac  Office,  for  reading  the  MS.,  and  to  Mr. 
I.  S.  Upson,  the  College  Librarian,  for  reading  the  proof  sheets. 

E.  A.  B, 
Rutgers  College,  ) 

New  Brunswick,  N.  J.,  May,  1890-  \ 


TABLE  OF  CONTENTS. 


INTRODUCTION. 

PAGE 

Definitions » 1 

Straight  Lines 4 

Plane  Angles 5 

Angular  Measure 8 

Superposition 10 

Postulates  and  Axioms 12 

Symbols  and  Abbreviations = o  13 


PLANE  GEOMETRY, 
BOOK  I. 

RECTILINEAR   FIGURES. 

Perpendicular  and  Oblique  Lines 14 

Parallel  Lines : , 23 

Triangles 33 

Quadrilaterals 46 

Polygons 52 

Miscellaneous  Theorems 56 

Symmetry 64 

Exercises. 67 

BOOK  11. 

THE   CIRCLE. 

Definitions 74 

Arcs  and  Chords 78 

Relative  Position  of  Two  Circles. ...   88 

Measurement  of  Angles 90 

Quadrilaterals , 102 

Problems  of  Construction. 104 

Exercises.    Theorems 123 

Loci 127 

Problems 129 

Y 


VI  TABLE  OF  C0NTENT8. 

BOOK  III. 

RATIO  AND  PROPORTION  :    SIMILAR  FIGURES. 

PAGE 

Ratio  and  Proportion 134 

Proportional  Lines 143 

Similar  Figures  149 

Similar  Triangles 150 

Similar  Polygons 155 

Numerical  Relations  between  the  Different  Parts  of  a  Triangle  158 

Problems  of  Construction 168 

Applications 173 

Exercises.    Theorems 175 

Problems 178 

BOOK  IV. 

AREAS   OF  POLYGONS. 

Measurement  of  Areas 18q 

Comparison  of  Areas  187 

Problems  of  Construction jgi 

Applications 200 

Exercises.    Theorems 201 

Problems 206 

BOOK  V. 

REGULAR  POLYGONS.      THE   CIRCLE.      MAXIMA  AND  MINIMA. 

Regular  Polygons 208 

The  Measure  of  the  Circle , 215 

Principle  of  Limits 216 

Problems  of  Construction 224 

Problems  of  Computation 229 

Value  of  Tt.     Method  of  Perimeters 231 

Maxima  and  Minima 234 

Exercises.    Theorems , 243 

Numerical  Exercises 245 

Problems _  .  247 

Exercises  in  Maxima  and  Miniina 248 


TABLE  OF  CONTENTS.  VU 

SOLID    GEOMETRY, 
BOOK  VI. 

PLANE  AND   SOLID  ANGLES. 

PAGE 

Definitions 250 

Lines  and  Planes 252 

Parallel  Planes 258 

Diedral  Angles.     Definitions 264 

Polyedral  Angles.     Definition'^ 276 

Exercises.    Theorems 282 

Loci. 284 

Problems 285 

BOOK  vn. 

rOLYEDRONS. 

Prisms  and  Parallelopipeds.     Definitions 286 

Pyramids 302 

Similar  Polyedrons 316 

Regnlar  Polyedrons 319 

General  Properties  of  Polyedrons 325 

Exercises.    Theorems 327 

Numerical  Exercises 329 

Problems 333 

BOOK  VIII. 

THE    SPHERE. 

Circles  of  the  Sphere  and  Tangent  Planes 334 

Spherical  Triangles  and  Polygons 342 

Relation  of  a  Spherical  Polygon  to  a  Polyedral  Angle  ...   344 

Symmetrical  Spherical  Triangles 345 

Polar  Triangles 349 

Relative  Areas  of  Spherical  Figures 354 

Exercises.   Theorems 360 

Numerical  Exercises 361 

Problems 363 

BOOK  IX. 

THE  THREE  ROUND  BODIES. 

The  Cylinder 366 

The  Cone 371 

The  Sphere 878 

Exercises.    Theorems. 386 

Numerical  Exercises 388 


OF  fhi 


ELEMENTARY  GEOMETRY. 


INTEODUCTION. 

Definitions. 

1.  Space  is  indefinite  extension  in  every  direction.  All 
material  bodies  occupy  limited  portions  of  space,  and  have 
length,  hreadth,  thick nes'i,  for m^  smd  positj^on.  The  mate- 
rial body  occupying  any  portion  of  space  is  called  a  physi- 
cal solid.  The  part  of  space  which  is  or  may  be  occupied 
by  a  material  body  is  called  a  geometric  solid.  A  physical 
solid  is  therefore  a  7-eal  body,  while  a  geometric  solid  is 
only  the  fonii  of  a  physical  solid,  and  is  the  one  treated  of 
in  Geometry.  The  term  solid  will  be  used  for  brevity  to  de- 
note a  geometric  solid. 

2.  A  solid  is  a  limited  portion  of  space,  and  has  length, 
hreadth,  and  thickness.  Length,  hreadth,  and  thickness  are 
called  the  three  dimensions  of  the  solid. 

3.  A  surface  is  the  limit  or  boundary  of  a  solid,  and  has 
only  two  dimensions,  length  and  breadth. 

A  surface  has  no  thickness,  for  if  it  had  any,  however  small,  it 
would  form  part  of  the  solid,  and  would  be  space  of  three  dimen- 
sions. 

4.  A  li7ie  is  the  limit  or  boundary  of  a  surface,  and  has 
only  one  dimension,  namely,  length. 

A  line  has  no  breadth,  for  if  it  had  any,  however  small,  it  would 
form  part  of  the  surface,  and  would  be  space  of  two  dimensions; 
and  if  in  addition  it  had  any  thickness,  it  would  be  space  of  three 
dimensions;  hence  a  line  has  neither  breadth  nor  thickness. 

1 


2  OEOMETRT. 

5.  A  point  is  the  limit  or  extremity  of  a  line,  and  has 
position,  but  neither  length,  breadth,  nor  thickness. 

A  point  has  no  length,  for  if  it  had  any,  however  small,  it  would 
form  part  of  the  line  of  which  it  is  the  extremity;  and  it  can  have 
neither  breadth  nor  thickness  because  the  line  has  none. 

6.  If  we  suppose  a  solid  to  be  divided  into  two  parts 
which  touch  each  other,  the  division  between  the  two  parts 
is  a  surface.  This  surface  can  have  no  thickness,  for  if  it 
had  a  thickness,  however  small,  it  would  be  a  part  either  of 
the  one  solid  or  the  other,  and  would  therefore  be  a  solid 
and  not  a  surface. 

Again,  if  we  suppose  a  surface  cut  into  two  parts  which 
touch  each  other,  the  division  between  the  two  parts  is  a 
line.  This  line  can  have  no  thickness,  because  the  surface 
has  none,  and  it  can  have  no  breadth,  for  it  forms  no  part 
of  either  surface. 

If  we  suppose  a  line  cut  into  two  parts  which  touch  each 
other,  the  division  between  the  two  parts  is  a  point.  This 
point  can  have  neither  breadth  nor  thickness,  because  the 
line  has  none,  and  it  can  have  no  length,  for  it  forms  no 
part  of  either  line. 

Euclid  regarded  a  point  merely  as  a  mark  of  position,  and  he  at- 
tached to  it  no  idea  of  size  and  shape. 

Similarly,  he  considered  that  the  properties  of  a  line  arise  only  from 
its  length  and  position,  without  reference  to  that  minute  breadth 
which  every  line  must  really  have  if  actually  drawn,  even  though 
the  most  perfect  instruments  are  used. 

We  cannot  make  the  points,  lines,  and  surfaces  of  Geometry.  A 
dot,  made  on  paper  or  on  the  blackboard,  will  have  length,  breadth, 
and  thickness,  and  hence  will  not  be  a  real  point.  Yet  the  dot  may 
be  taken  as  an  imperfect  representation  of  the  real  point.  So  also  a 
line,  drawn  on  paper  or  on  the  blackboard,  will  have  breadth  and 
thickness,  and  hence  will  not  be  a  real  line.  Yet  the  line  which  we 
draw  may  be  taken  to  represent  the  real  line. 

7.  We  have  considered  a  surface  as  the  boundary  of  a, 
solid,  a  line  as  the  boundary  of  a  surface,  and  a  point  as  the 
limit  of  a  line.     On  the  other  hand,  inversely,  we  may  re- 


INTRODUCTION'.  3 

gard  a  line  as  generated  by  the  motion  of  a  point,  a  surface 
as  generated  by  the  motion  of  a  line,  and  a  solid  as  generated 
by  the  motion  of  a  surface.  Again,  each  of  these  may  be  re- 
garded in  a  purely  abstract  manner,  distinct  from  each  other. 

Thus,  we  may  suppose  a  surface  to  exist  in  space  sepa- 
rately from  the  solid  whose  boundary  it  forms,  and  to  be  of 
iDilimited  extent. 

Similarly,  we  may  suppose  a  line  to  exist  in  space  sepa- 
rately from  the  surface  whose  boundary  it  forms,  and  to  be 
of  tinlimited  lengtli. 

Likewise  we  may  suppose  a  point  to  exist  in  space  sepa- 
rately from  the  line,  and  to  have  only  position. 

The  points,  lines,  surfaces,  and  solids  of  Geometry  are 
called  geometric  points,  lines,  surfaces,  and  solids. 

8.  A  draight  line,  or  rigid  line,  is 

one  which  has  the  same  direction  at    A         -  B 

every  point,  as  the  line  AB. 

9.  A  curved  line  is  one    no  part  

of  which  is  straight,  but  changes  its     C^^^^'  D 

direction  at  every  point,  as  the  line 

CD. 

10.  A  broken  line  is  a  line  made 


up   of    different  successive    straight     ^  p.    , 

lines,  as  the  line  EF. 


The  word  line,  used  alone,  signifies  a  straight  line;  and 
the  word  curve,  a  curved  line. 

11.  K  plane  surface,  or,  simply,  a  plane,  is  a  surface  in 
which  the  right  line  joining  any  two  points  in  it  lies  wholly 
in  the  surface. 

12.  A  curved  surface  is  one  no  part  of  which  is  plane. 

13.  A  figure  is  any  definite  combination  of  points,  lines, 
surfaces,  or  solids. 

A  plane  figure  is  one  formed  of  points  and  lines  in  a 
plane.  If  the  figure  is  formed  of  right  lines  only,  it  is 
called  a  rectilinear,  or  right-lined,  figure. 

The  figure  of  a  solid  depends  upon  the  relative  position 


4  GEOMETRY. 

of  the  points  in  its  surface.  Lines,  surfaces,  and  solids  are 
the  geometric  figures.  When  the  extent  of  lines,  surfaces, 
and  solids  is  considered  they  are  called  magnitudes,  but  when 
their /o/-m  or  sJiape  is  considered  they  are  csiWed  figures. 

14.  Geometry  is  the  science  which  treats  of  magnitude, 
form,  and  position.  Thus  it  is  the  province  of  Geometry 
to  investigate  the  properties  of  solids,  of  surfaces,  and  of 
the  figures  constructed  on  surfaces. 

Plane  Geometry  treats  of  plane  figures. 

Solid  Geometry,  called  also  Geometry  of  Space  and  Geom- 
etry of  TJiree  Dimensions,  treats  of  solids,  of  curved  sur- 
faces, and  of  the  figures  described  on  curved  surfaces. 

Straight  Lines. 

15.  K  finite  straight  line  is  a  straight  line  contained  be- 
tween two  definite  points  which  are  its  extremities.  When 
a  straight  line  is  produced  indefinitely  it  is  called  an  in- 
definite straight  line.  Any  finite  straight  line  may  be  sup- 
posed at  anytime  to  be  produced  into  an  indefinite  straight 
line. 

Two  finite  straight  lines  are  said  to  be  equal,  or  of  equal 
length,  when  the  extremities  of  the  one  line  can  be  made  to 
coincide  respectively  with  the  extremities  of  the  other. 

If  any  line,   as  OB,   be   produced 

through  0  to  A,  the  parts  OB  and  OA    A^ -^ B 

are  said  to  have  opposite  dii'ections  Fig.  2 

from  the  common  point  0. 

Every  straight  line  AB  has  two  opposite  directions,  the 
one  from  A  toward  B,  expressed  by  *'  the  line  AB,''  and  the 
other  from  B  toward  A,  expressed  by  "the  line  BA.'' 

If  a  line  BC  is  to  be  produced  toward  I),  we  should  ex- 
press this  by  saying  that  "  BC  is  to  _ 
be  produced  "\  but  if  it  is  to  be  pro-               B        C 
duced  toward  A,  we  should  express  ^*^'  ^ 
this  by  saying  that  '^  CB  is  to  be  produced." 

Straight  lines  are  added  together   by  ])lacing  them   one 


INTRODUCTION.  O 

after  anotlier  in  succession  in  the  same  straight  line  so  that 
one  extremity  of  each  newly  added  line  coincides  with  one 
extremity  of  the  last  added  line,  and  so  that  no  part  of  any 
newly  added  line  coincides  with  any  part  of  the  last  added 
line. 

Thus,  AB,  BO,  and  CD,  Fig.  3,  are  added  together  and 
form  the  straight  line  AD. 

AB,  BO,  and  CD  are  called  the  jmrts  of  AD,  and  AD  is 
called  the  sum  of  AB,  BC,  and  CD. 

Plane  Akgles. 

16.  K  plane  ancjle  is  the  opening  between  two  straight 
lines  drawn   from   the   same  point.  ^B 
The  straight  lines  are  called  the  arms 
or  sides  of  the  angle,  and  the  common 

point  is  called  its  vertex.     Thus  the    q^ A 

lines  OA,  OB  are  said  to  contain,  or  ^'9'  * 

include^  or  form  the  angle  at  0. 

When  there  are  several  angles  at  one  point,  any  one  of 
them  is  expressed  by  three  letters,  putting  the  letter  at 
the  vertex  between  the  other  two. 
Thus,  if  the  straight  lines  OA,  OB, 
OC  meet  at  the  point  0,  the  angle 
contained  by  the  lines  OA,  OB  is 
named  the  angle  AOB  or  BOA;  the 
angle  contained  by  the  lines  OA, 
OC  is  named  the  angle  AOC  or  COA;    '  '"'9-  ^ 

and  the  angle  contained  by  OB,  OC  is  named  the  angle 
BOO  or  COB. 

When  there  is  only  one  angle  at  a  point,  it  may  be  de- 
noted either  by  the  single  letter  at  that  point,  or  by  three 
letters  as  above.  Thus  in  Fig.  4  the  angle  at  the  point 
0  may  be  denoted  either  by  the  angle  0  or  by  AOB  or  by 
BOA. 

17.  Adjacent  angles  are  angles  which  have  a  common 
vertex  and  one  common  arm,  their  non-coincident  arms 


6 


GEOMETRY. 


being  on  opposite  sides  of  the  common  arm.  Thus  the 
angles  A  OB  and  COB  (Fig.  5)  are  adjacent  angles,  of  which 
OB  is  the  common  arm. 

Of  the  two  straight  lines  OB,  00  (Fig.  5)  it  is  easily 
seen  that  the  opening  between  OA  and  OC  is  greater 
than  the  opening  between  OA  and  OB.  This  we  express 
by  saying  that  the  angle  AOC  is  greater  than  the  angle 
AOB. 

The  7nagnitude  or  size  of  an  angle  depends  entirely  upon 
the  extent  of  opening  between  its  sides,  and  is  not  altered 
by  changing  the  length  of  its  sides. 

18.  Angles  are  equal  when  they  can  be  placed  one  upon 
the  other  so  that  the  vertex  and  sides  of 

the  one  can  be  made  to  coincide  with  the 
vertex  and  sides  of  the  other.  Thus 
the  angles  ABO  and  DEF  are  equal 
if  ABO  can  be  placed  upon  DEF  so 
that  while  BA  coincides  with  ED,  BO 
shall  also  coincide  with  EF. 

19.  The  angle  formed  by  joining  two 
or  more  angles  together  is  called  their 
sum.     Angles  are  added  together  by  placing  them  so  as  to 
be  adjacent  to  each  other.     Thus 
the  sum  of  the  two  angles  ABC, 
PQK,  is  the  angle  ABE,  formed 
by  applying  the  side  QP  to  the 
side   BO   so  that   the   vertex  Q 
shall  fall  on  the  vertex  B,  and 
the  side  QR  on  the  opposite  side 
of  BO  from  BA. 

If  the  angles  ABC,  PQR  are 
equal  to  each  other,  the  angle  ABR  is  doiihle  either  of 
them,  and  the  common  side  BC  is  said  to  hisect  the  angle 
between  the  non-coincident  sides  BA  and  BR. 

20.  When  a  straight  line  standing  on  another  makes  the 
adjacent  angles  equal  to  each  other,  each  of  the  angles  is 


Fig.  6 


INTRODUCTION. 


0 
Fig.  8 


called  a  right  angle;  and  the  straight  line  which  stands  on 
the  other  is  said  to  be  perpendicu- 
lar or  at  right  angles  to  it.  Thus,  if 
the  adjacent  angles  AOC  and  BOO 
are  equal  to  each  other,  each  is  a 
right  angle,  and  the  line  CO  is  per- 
pendicular to  AB.  The  point  0  is 
called  the  foot  of  the  perpendicular. 

21.  A  straight  angle  has  its  arms  extending  in  opposite 
directions  so  as  to  be  in  the  same 
straight  line.  Thus,  if  the  arms 
OA,  OB  are  in  the  same  straight 
line,  the  angle  formed  by  them  is 
called  a  straight  angle. 

Since  the  sum  of  the  two  right 
angles  AOC  and  BOC  (Fig.  8)  is  the  angle  AOB  (19)*,  a 
right  angle  is  half  a  straight  angle. 

22.  An  acute  angle  is  an  angle 
which  is  less  than  a  right  angle,  as  ^^ 
the  angle  A. 

23.  An  obtuse  angle  is  an  angle  which  is  greater  than  a 
right  angle,  and  less  than  a  straight 
angle,  as  the  angle  BAC. 

24.  When  the  sum  of  two  -angles 
is  a  right  angle,  each  is  called  the  A  ^ 
complement  of  the   other,  and  the                 ^'9-  " 

two  are  called  complementary  angles.     Thus,  if  the  angle 
BAG  is  a  right  angle,  the  angles  BAD, 
J)AC  are  complements  of  each  other. 

25.  When  the  sura  of  two  angles  is  a 
straight  angle,  each  is  called  the  supplement 
of  the  other,  and  the  two  are  called  sup- 
plementary  adjacent  angles.  Thus,  if  the 
angle  AOB  (Fig.  9)   is  a  straight  angle, 


Fig.  12 


the  angles  BOC,  CO  A  are  supplements  of  each  other. 


*  An  Arabic  uumenil  in  parenthesis  refers  to  an  article. 


8 


OEOMETBY. 


Hence,  when  one  line  stands  on  another,  the  two  adjacent 
angles  are  supplements  of  each  other.  Hence  a  right  angle 
is  equal  to  its  supplement. 

The  supplement  of  an  acute  angle  is  obtuse,  and,  con- 
versely, the  supplement  of  an  obtuse  angle  is  acute. 

26.  A  reflex  angle  is  an  angle  which  is  greater   than 
a   straight  angle,  and   less  than   two 
straight  angles,  as  the  angle  0. 

Acute,  obtuse,  and  reflex  angles  are 
called  oblique  angles,  in  distinction 
from  right  and  straight  angles ;  and 
intersecting  lines  which  are  not  per- 
pendicular to  each  other  are  called  oblique  lines. 

27.  Where  two  angles  are  contained  between  two  inter- 
secting lines  on  opposite  sides  of  the  A>^  ^D 
vertex,  they  are  called  opposite  or  ver- 
tical angles.     Thus,  AOC  and  BOD 
are  6pposite  or  vertical  angles,  as  also 
AOD  and  COB. 


Fig.  14 


An^gular  Measure. 


28.  A  right  line  drawn  from  the  vertex  and  turning 
about  it  in  the  plane  of  the  angle  from  the  position  of  co- 
incidence with  one  side  of  the  angle  to  that  of  coincidence 
with  the  other  side,  is  said  to  turn  through  the  angle,  and 
the  angle  is  the  greater  as  the  quantity  of  turning  is  greater. 

Thus,  suppose  that  the  right  line 
OP  (Fig,  15)  is  capable  of  revolv- 
ing about  the  point  0,  like  the 
hands  of  a  watch,  but  in  the  oppo- 
site direction,  and  that  it  has 
passed  successively  from  the  po- 
sition OA  to  the  positions  occupied 
by  OB,  00,  OE,  etc.  Then  it  is 
clear  that  the  line  must  have  done  F"'g.  's 

more  turning  in  passing  from  OA  to  00  than  in  passing 


INTRODUCTION.  9 

from  OA  to  OB;  and  consequently  the  angle  AOC  is  said  to 
be  greater  than  the  angle  AOB. 

When  the  revolving  line  OP  turns  from  coincidence  with 
OA  to  the  position  OB  it  is  said  to  describe  or  to  generate 
the  angle  AOB.  When  the  revolving  line  has  turned  from 
the  position  OA  to  the  position  OD,  perpendicular  to  OA, 
it  has  generated  the  right  angle  AOD  ;  when  it  has  turned 
to  the  position  OA',  it  has  generated  the  straight  angle 
AOA';  when  it  has  turned  to  the  position  OF,  it  has  gen- 
erated the  reflex  angle  AOF;  when  it  has  turned  entirely 
around  to  the  position  OA,  it  has  generated  two  straight 
angles. 

Hence,  the  whole  angle  which  a  line  must  turn  through, 
about  a  point  in  a  plane,  to  take  it  around  to  its  first 
position,  is  two  straight  angles,  or  four  right  angles. 

Again,  since  the  revolving  line  may  turn  from  one  po- 
sition to  the  other  in  either  of  two   b^ 
directions,  two  angles  are  formed  by 
two  lines  drawn  from  a  point. 

Thus,  if  OA,  OB  be  the  sides  of  an 
angle,  a  line  may  turn  from  the  po- 
sition OA  to  the  position  OB  about 
the  point  0  in  either  of  the  two  directions  indicated  by  the 
arrows,  giving  the  obtuse  angle  AOB  (marked  a),  and  the 
reflex  angle  AOB  (marked  h). 

Angles  are  generally  measured  in  degrees,  minutes,  and 
seconds.  A  degree  is  the  ninetieth  part  of  a  right  angle,  or 
the  three  hundred  and  sixtieth  part  of  four  right  angles. 
A  minute  is  the  sixtieth  part  of  a  degree;  and  a  second  is 
the  sixtieth  part  of  a  minute.  Degi-ees,  minutes,  and 
seconds  are  denoted  by  the  symbols  °,  ',  ".  Thus  7 
degrees,  24  minutes,  and  38  seconds  is  written,  7°  24'  38". 

Hence,  when  the  revolving  line  OP  (Fig.  15)  has  turned 
through  one-fourth  of  a  revolution,  it  has  generated  a  right 
angle,  or  90°  ;  when  it  has  made  half  a  revolution,  it  has 
generated  a  straight  angle,  or  180° ;  and  when  it  has  made 


\ 


10  GEOMETRY. 

a  luliole  revolution,  it  has  generated  two  straight  angles,  or 

360°. 

Superposition. 

29.  The  placing  of  one  geometric  magnitude  on  another, 
such  as  a  line  on  a  line,  an  angle  on  an  angle,  etc.,  is 
called  superposition.  The  superposition  employed  in 
Geometry  is  only  mental,  that  is,  we  conceive  one  magni- 
tude to  he  taken  up  and  laid  down  upon  the  other;  and 
then,  if  we  can  prove  that  they  coincide,  we  infer  that 
they  are  equal.  This  is  the  ultimate  test  of  the  equality  of 
two  geometric  magnitudes. 

Thus,  if  two  straight  lines  are  to  be  compared,  we  con- 
ceive one  to  be  taken  up  and  placed  on  the  other,  and  find 
whether  their  two  ends  can  be  made  to  coincide.  If  so, 
they  are  equal  ;  if  not,  they  are  unequal.  If  two  angles 
are  to  be  compared,  we  conceive  one  to  be  taken  up  and 
applied  to  the  other.  If  they  can  be  so  placed  that  their 
vertices  coincide  in  position  and  their  sides  in  direction, 
the  angles  are  equal  (18). 

Superposition  involves  the  principle*  that  '*any  figure 
may  be  taken  up,  transferred  from  one  position  to  another, 
and  laid  down  again  without  change  of  form  or  size." 

Magnitudes  which  coincide  with  one  another  throughout 
their  whole  extent  are  said  to  be  equal. 

Definitions  of  Terms. 

30.  A  theorem  is  a  truth  which  requires  proof. 

31.  A  prohlem  is  a  question  which  requires  solution, 
such  as  a  particular  line  to  be  drawn,  or  a  required  figure 
to  be  constructed. 

32.  An  axiom  is  a  self-evident  truth,  which  is  admitted 
without  proof. 

33.  K  post-date  assumes  the  possibility  of  solving  a  cer- 
tain problem. 

*  Euclid  makes  frequeiit  use  of  this  principle,  witJiQut  explicitly 
gtatin^  it. — Casey. 


INTRODUCTION,  11 

34.  A  jn'oposition  is  a  general  term  for  a  theorem, 
problem,  axiom,  or  postulate. 

35.  A  demonstration  is  the  course  of  reasoning  by  which 
we  prove  a  theorem  to  be  true. 

36.  A  corollary  is  a  conclusion  which  folloAvs  imme- 
diately from  a  theorem. 

37.  A  lemma  is  an  auxiliary  theorem  required  in  the 
demonstration  of  a  principal  theorem. 

38.  A  scJiolium  is  a  remark  upon  one  or  more  propo- 
sitions. 

39.  An  hypothesis  is  a  supposition  made  either  in  the 
enunciation  of  a  proposition  or  in  the  course  of  a  demon- 
stration. 

40.  A  solution  of  a  problem  is  the  method  of  construc- 
tion which  accomplishes  the  required  end. 

41.  A  construction  is  the  drawing  of  such  lines  and 
curves  as  may  be  required  to  prove  the  truth  of  a  theorem, 
or  to  solve  a  problem. 

42.  The  enunciation  of  a  theorem  consists  of  two  parts  : 
the  hypothesis,  or  that  which  is  assumed ;  and  the  conclusion, 
or  that  which  is  asserted  to  follow  therefrom. 

Thus,  in  the  typical  theorem. 

If  A  is  B,  then  C  is  D, 

the  hypothesis  is  that  A  is  B,  and  the  conclusion  that  0 
is  D. 

43.  The  emcnciafion  of  a  problem  consists  of  two  parts: 
the  data,  or  things  supposed  to  be  given;  and  the  qiimsita, 
or  things  required  to  be  done. 

44.  Two  theorems  are  said  to  be  converse,  each  of  the 
other,  whenj;he  hypothesis  of  each  is  the  conclusion  of  the 
other. 

Thus,  If  C  is  D,  then  A  is  B, 

is  the  converse  of  the  typical  theorem  in  (42). 


12  GEOMETBT, 


Postulates. 

45.  Let  it  be  granted — 

1.  That  a  straight  line  may  be  drawn  from  any  one  point 
to  any  oth'er  point. 

2.  That  a  terminated  straight  line  may  be  produced  to 
any  distance  in  a  straight  line. 

3.  That  a  circle  may  be  described  with  any  centre,  at 
any  distance  from  that  centre. 

Axioms. 

46.  1.  Things  which  are  equal  to  the  same  thing  are 
equal  to  each  other. 

2.  If  equals  be  added  to  equals  the  sums  will  be  equal. 

3.  If  equals  be  taken  from  equals  the  remainders  will  be 
equal. 

4.  If  equals  be  added  to  unequals  the  sums  will  be  un- 
equal. 

5.  If  equals  be  taken  from  unequals  the  remainders  will 
be  unequal. 

6.  Things  which  are  double  the  same  thing,  or  equal 
things,  are  equal  to  one  another. 

7.  Things  which  are  halves  of  the  same  thing,  or  of  equal 
things,  are  equal  to  one  another. 

8.  The  whole  is  greater  than  any  of  its  parts. 

9.  The  whole  is  equal  to  the  sum  of  all  its  parts. 

Geometric  Axioms. 

10.  A  straight  line  is  the  shortest  distance  between  any 
two  points. 

11.  If  two  straight  lines  have  two  points  in  comm  on,  they 
will  coincide  throughout  their  whole  length,  and  form  but 
one  straight  line. 

12.  Through  a  given  point  only  one  straight  line  can  be 
drawn  parallel  to  a  given  straight  line. 


INTRODUCTION. 


13 


Symbols  and  Abbreviations. 
47.  The  following  are  some  of  the  principal  symbols  and 
abbreviations  that  are  commonly  used  in  Colleges  and 
Public  Schools.  These  symbols  and  abbreviations  are  gen- 
erally employed  in  the  present  book.  They  are  recom- 
mended to  the  student  for  writing  out  the  propositions  and 
demonstrations  on  the  blackboard  or  in  exercise  books,  as 
their  use  will  greatly  shorten  the  work. 

-|-  plus,  or  together  with.  adj adjacent. 

—  minus,  or  diminished  by.  alt alternate. 

X  multiplied  by.  ax axiom. 

-^  divided  by.  cons construction. 

.'.  therefore.  cor corollary. 

_   j  equals,  cyl cylinder. 

~~    (  or  is  (or  are)  equal  to.  def definition. 

>  is  (or  are)  greater  than.  ext exterior. 

<  is  (or  are)  less  than.  fig figure. 

Z  angle.  hyp hypothesis. 

Zs  angles.  int interior. 

A  triangle.  opp opposite. 

As  triangles.  post postulate. 

£17  parallelogram.  prob problem. 

ZZ7s  parallelograms.  prop proposition. 

±   perpendicular.  pt point. 

±s  perpendiculars.  rt right. 

1 1   parallel.  sim similar. 

I  Is  parallels.  sq square. 

O  circle.  st straight. 

Os  circles.  sup supplementary. 

Oce  circumference. 
0ces  circumferences. 

The  words  ''join  AB^'  are  used  as  an  abbreviation  for 
*'draw  a  straight  line  from  A  to  B.^' 

The  initial  letters  Q.  e.  d.,  placed  at  the  end  of  a  theorem, 
stand  for  the  Latin  words  duod  erat  Demonstrandum, 
meaning  loliich  tvas  to  be  proved. 

The  letters  Q.  E.  F.,  placed  at  the  end  of  a  problem,  stand 
for  duod  erat  Faciendum,  which  was  to  be  done. 


PLANE  GEOMETRY. 


Book  I. 
EECTILINEAR  FIGURES. 


Perpen^dicular  and  Oblique   Lines. 
Proposition  1 .    Theorem. 

48.  All  straight  angles  are  equal  to  one  another.* 
Hyp.  Let  AB,  AC  be  the  arms 

of  a  st.  Z.   whose  vertex  is  A,  and     ^  J  'C 

DE,  DF  the  arms  of  another  st.  Z 

whose  vertex  is  D.  ^  ^  ^ 

To  prove        /  BAG  =  Z  EDF. 

Proof.     Because  the  Z  BAG  is  a  st.  Z  ^ 

.'.  BA  and  AG  are  in  the  same  st.  line  BAG.       (21) 

Because  the  Z  EDF  is  a  st.  Z , 

.-.  ED  and  DF  are  in  the  same  st.  line  EDF.       (21) 

Kow  if  the  Z  BAG  be  applied  to  the  Z  EDF  so  that  the 
vertex  A  shall  coincide  with  the  vertex  D  and  the  arm  AB 
with  the  arm  DE,  then  the  arm  AG  will  coincide  with  the 
arm  DF, 

because,  if  two  st.  lines  have  tivo  p)oints  in  common  they 
will  coincide  throughoiit  their  ivhole  lengthy  and  form  hut 
one  st.  line  (Ax.  11). 

.-.  Z  BAG  =  Z  EDF.  q.e.d. 

49.  Cor.  1.  All  right  angles  are  equal  to  one  another.  \ 
(21),  also  (Ax.  7). 

50.  Cor.  2.  The  complements  of  equal  angles  are  equal 
to  each  other,  and  the  siq^plemeiits  of  equal  angles  are  equal 
to  each  other. 

51.  Cor.  ^.  At  a  given  j^oint  in  a  given  straight  line 
only  one  perpendicular  can  he  drawn  to  that  line. 

*  The  Elements  of  Plane  Geometry,  by  Association  for  the  Improvement  of 
Geometric  Teaching,  p,  20a. 
t  This  is  Euclid's  11th  axiom. 

14 


BOOK  L— PERPENDICULAR  AND  OBLiqUE  LINES.  15 


Proposition  2.    Theorem. 

52.  If  one  sir ai gilt  line  meet  another  straight  line,  the 
sum  of  the  two  adjacent  angles  ii  equal  to  two  right  angles. 

Hyp.  Let  the  st.  line  DO  meet  the 
st.  line  AB  at  C. 

To  prove 

Z  ACD  +  Z  DOB  =  2  rt.  Zs.  ,^ 

Proof  From  C  draw  CE  at  rt.  Zs     ^ c  ^ 

to  AB. 

Then  Z  ACD  =  Z  ACE  +  Z  ECD. 

Add  Z  DCB  to  each. 

.-.  Z  ACD+  Z  DCB=  Z  ACE+  z  ECD  +  Z  DCB. 

ijf  equals  be  added  to  equals,  the  sums  will  be  equal  {Ax.  3). 

Again,  Z  ECB  ^  Z  ECD  +  Z  DCB. 

Add  Z  ACE  to  each. 

.-.  Z  ACE+  Z  ECB=  Z  ACE+  Z  ECD  +  Z  DCB. 
If  equals  be  added  to  equals,  the  sums  will  be  equal  {Ax.  2). 

Hence 

Z  ACE+  Z  ECB  =r  z  ACD  +  Z  DCB.  (Ax.  1)* 

But  Z  ACE  +  Z  ECB  :=  2  rt.  Z  s.      .  (Cons. ) 

.-.  Z  ACD+ zDCB  =  2rt.Zs.  q.e.d. 

ScH.  The  angles  ACD,  DCB  are  supplementary  adjacent 
angles  (25). 

53.  Cor.  If  one  of  the  angles  ACD,  DCB  is  a  right 
angle,  the  other  is  also  a  right  angle. 

The  sum  of  all  the  angles  on  the  same  side  of  a  straight 
li7ie,  at  a  cdmbmo^i  point,  is  equal  to  two  right  angles  ;  for 
their  sum  is  equal  to  the  sum  of  the  two  adjacent  angles 
ACD,  DCB. 

♦  Tbe  student  should  quote  every  reference  in  full. 


W  PLANE  GEOMETRY. 


Proposition  3.    Theorem. 

54.  Conversely,  if  the  sum  of  two  adjacent  angles  is  equal 
to  two  right  angles,  their  exterior  sides  are  in  oiw  straight 
line. 


-B 


Hyp.  Let  Z  ACD  +  Z  DCB  =  2rt.  Zs. 

To  prove  that  AC  and  CB  are  in  one  st.  line. 

Proof.  If  CB  be  not  in  the  same  st.  line  as  AC,  let  CE 
be  in  the  same  st.  line  as  AC. 

ThenZ  ACD  +  Z  DCE  =  2  rt.  Zs. 
being  sup.  adj.  I  s  (52). 

But    z  ACD+  Z  DCB  =  2i»t.  Zs.  (Hyp.) 

.-.  Z  ACD  +  Z  DCE  =  Z  ACD  +  Z  DCB.  (Ax.  1) 

Take  away  the  common  Z  ACD. 

.-.  Z  DCE=  ZDCB,  (Ax.  3) 

which  is  impossible  (Ax.  8),  unless  CE  coincides  with  CB. 

.'.  AC  and  CB  are  in  one  st.  line.  Q.E.D. 


EXERCISES. 


1.  Find  the  number  of  degrees  in  an  angle  if  it  is  three 
times  its  complement. 

2.  Find  the  number  of  degrees  in  an  angle  if  its  com- 
plement and  supplement  are  together  equal  to  U0°.   -^  ^  " 


BOOK  I.—PEUPENI)ICULAU  AND  OBLiqUE  LINES.   17 


Proposition  4-    Tlieorem. 

55.  If  two  atraiglit  lines  mtersect  each  other,  the  vertical 
angles  are  equal. 


-B 


Hyp,  Let  the  st.  lines  AB  and  CD  cut  at  E. 
To  prove  Z  AEC  =  Z  BED. 

Proof.    Z  CEA  +  Z  CEB  =  2  rt.  Zs, 
being  sup.  adj.  Z«  (52). 
Z  AEC  +  Z  AED  =  2  rt.  Zs, 
being  sup.  adj .  Z«  (52). 

.-.  Z  CEA  +  Z  CEB  =  z  AEC  +  Z  AED. 

^^^  rt.  Z«  are  egw«^  to  one  another  (49). 

Take  away  the  common  Z  AEC. 

.-.  Z  CEB  =  Z  AED.  (Ax.  3) 

In  the  same  way  it  may  be  proved  that 

Z  AEC  =  Z  BED.  Q.E.D. 

56.  Cor.  1.  If  two  straight  lines  intersect  each  other, 
the  four  angles  which  they  make  at  the  point  of  intersection, 
are  together  equal  to  four  right  angles. 

If  one  of  the  four  angles  is  a  right  angle,  the  other  three 
are  right  angles,  and  the  lines  are  mutually  perpendicular 
to  each  other. 

57.  Cor.  2.  If  any  numher  of  straight  lines  meet  at  a 
point,  the  sjim  of  all  the  angles  made  by  consecutive  lines,  in 
a  plane,  is  equal  to  four  right  angles. 


18 


PLANE  GEOMETRY. 


I 


Proposition  5.    Theorem. 

58.  From  a  i^oint  without  a  straight  line  only  one 'per- 
pendicular can  he  drawn  to  that  line;  and  this  perpen- 
dicular is  the  shortest  distance  from  the  point  to  the  line. 

Hyp,  Let  AB  be  the  given  st, 
line,  P  the  point  without  the  line, 
PC  a  ±  from  P  to  AB,  and  PD 
any  other  line  from  P  to  AB. 

(1)  To  prove  that  PD  is  not  ± 
to  AB. 

Proof.  Let  the  part  of  the  plane 
above  the  line  AB,  and  which  con- 
tains P,  be  revolved  about  AB  as  \' 
an  axis  until  the  point  P  comes  P 
into  the  position  P'. 

Then,  since  Z  ACP  is  a  rt.  Z ,  Z  ACP'  is  also  a  rt.  Z .  (29) 
.-.  PCP'  is  a  St.  line.  (54) 

If  Z  ADP  is  a  rt.  Z ,  Z  ADP'  is  also  a  rt.  Z .  (29) 

.-.  PDP'  is  a  St.  line;  (54) 

that  is,  between  two  points  there  are  two  straight  lines, 

which  is  impossible.  (Ax.  11) 

.-.  Z  ADP  is  not  a  rt.  Z ,  and  .-.  PD  is  not  ±  to  AB.  (20) 

.*.  PC  is  the  only  ±  to  AB  from  the  point  P. 

(2)  To  prove  PC  <  PD. 

Proof.  Since  in  the  revolution  of  CPD  about  AB  as  an 
axis,  the  point  P  comes  into  the  position  P', 
.-.  PC  =  P'C,  and  PD  =  P'D; 
.-.  PC  +  P'C  =  2  PC, 
and  PD  +  DP'  =  2  PD. 

PC  +  P'C  <  PD  +  DP'. 
.•.2PC<2PD,  and.-.  PC  <  PD. 


But 


(29) 

(Ax.  2) 

(Ax.  2) 

(Ax.  10) 


Q.E.D. 

59.  ScH.  By  the  distance  of  a  point  from  a  line  is  meant 
the  shortest  distance,  that  is,  the  length  of  the  perpendicu- 
lar from  the  point  to  the  line. 


BOOK  I.-PERPENDICULAR  AND  OBLIQUE  LINES.   19 


Proposition  6.    Theorem. 

60.  Two  oblique  lines  drawn  from  a  j)oint  to  a  straight 
line,  cutting  off  equal  distances  from  tlie  foot  of  the  perpen- 
dicular, are  equaL 


Hyp.  Let  CD  be  the  ±  from  C  to  the  line  EF,  and  CA 
and  CB  two  oblique  lines  so  that  AD  =  DB. 

Tojirove  CA  =  CB. 

Proof,  Let  the  part  CAD  be  revolved  about  CD  as  an 
axis  until  it  comes  into  the  plane  of  CDB. 


Because  Z  CDA  ^  Z  CDB  =  rt.  Z , 

.*.  DE  will  take  the  direction  of  DF. 
Because  DA  =  DB, 

the  point  A  will  fall  upon  the  point  B. 
.•.CA  =  CB. 


(Hyp.) 
(Hyp.) 


(Ax.  11) 

Q.E.D. 


61.  Cqr.  Two  equal  ohlique  lines  draion  from  a  point 
to  a  straight  line,  make  equal  angles  with  that  line,  and  cut 
off  equal  distances  from  the  foot  of  the  perpendicular^ 


20  PLANE  GEOMETRY. 


Proposition  7.    Theorem. 

62.  If  from  any  point  two  lines  he  drawn  to  the  ends  of 
a  straight  line,  their  sum  will  he  greater  than  the  sum  of 
tiuo  other  lines  similarly  draion,  hut  included  hy  them. 


'  Hyp.  Let  AB,  AC  be  two  lines  drawn  from  the  point  A 
to  the  ends  of  the  line  BC,  and  DB,  DC  two  lines  similarly 
drawn,  but  included  by  AB,  AC. 

To  prove  AB  +  AC  >  DB  +  DC. 

Proof     Produce  BD  to  meet  AC  at  E. 

Then  BA  +  AE  >  BE.  (Ax.  10) 
Add  EC  to  each. 

.-.  BA  +  AC  >  BE  +  EC.  (Ax.  2) 

Again,  DE  +  EC>  DC.  (Ax.  10) 
Add  BD  to  each. 

.  • .  BE  +  EC  >  BD  +  DC.  (Ax.  2) 

Much  more  .  • .  BA  +  AC  >  BD  +  DC.  q.e.d. 

EXERCISES. 

1.  Write   out   in  full  the  proof  of   the  second  part  of 
Prop.  4,  that  the  angle  AEC  is  equal  to  the  angle  BED. 

2.  Prove  that  the  straight  line  which  bisects  the  angle 
AEC,  in  Prop.  4,  bisects  also  the  vertical  angle  BED. 


BOOK  L-PERPENDICULAB  AND  OBLIQUE  LINES.  21 

Proposition  8.    Theorem. 

63.  Of  two  oblique  lines  draiun  from  the  same  point  to 
the  sa7ne  straight  line,  that  lohich  meets  the  line  at  the 
greater  distance  from  the  foot  of  the  perpendicular  is  the 

greater. 

P 


Hyp.  Let  PC  be  _L  from  P  to  AB,  and  PE,  PD  two 
oblique  lines  so  that  CE  >  CD. 
Toj^rove  PE  >  PD. 

Proof  Produce  PC  to  P',  making  CP'  =  PC. 

Join  DP',  EP'. 
Because  AB  is  ±  to  PP'  at  its  middle  point, 

.-.  PD  :=  DP',  and  PE  ==  EP'.  (60) 

But  PE  +  EP'  >  PD  +  DP'.  (62) 

.•.2PE>2PD,  .-.  PE>PD. 
If  tliQ  two  oblique  lines  are  on  opposite  sides  of  PC,  as 
PE  and  PD',  and  if  CE  >  CD',  take  CD  =  CD',  and  join 
PD. 

Then  PD  =  PD'.  (60) 

But  PE  >  PD,  as  just  proved. 

.-.  PE  >   PD'.  Q.E.D. 

64.  CoR.  1.  Only  two  equal  straiglit  lines  can  he  draion 
from  a  point- to  a  straight  line, 

65.  Cor.  2.  Of  two  unequal  oblique  lines,  the  greater 
exits  off  the  greater  distance  from  the  foot  of  the  perpen- 
dicnlar. 


22 


PLANE  GEOMETRY. 


Proposition  9.    Tiieorem. 

1^      66.  1.  Every  ijoint  m  the  perpendicular  erected  at  the 
middle  of  a  straight  line  is  equally  distant  from  the  e.c- 
t7'emities  of  the  line. 
\      2.  Every  point  without  the  perpendicular  is  unequal'ty 
\  distant  from  the  extremities  of  the  line. 
Hyp.     Let  DO  be  1  to  AB  at  its 
middle  point  G,  P  any  point  in  DO, 
and  0  any  point  without  DO. 
Draw  PA,  PB,  and  OA,  OB. 

(1)  To  prove  PA  =  PB. 
Proof  Because  DO  1  to  AB,  and  AO  =  OB, 

.  • .  PA  =  PB. 

(2)  To  prove  OA  >  OB. 
Proof.  OA  will  cut  the  1  DO  at  E;  join  EB. 
Then                       OB  <  OE  +  EB. 
But  EB  =  EA. 

.-.  OB<OE  +  EA, 

or  OB  <  OA. 

67.  Cor.  Every  point  equally  distant  from  the  extremities 

of  a  straight  line  lies  in  the  perpendicular  hisector  of  the  line. 

If  a  straight  line  have  tiuo  points,  each  ofivhich  is  equally 

\distant  from  the  extremities  of  a  second  line,  it  will  be  per- 

lendicular  to  the  second  line  at  its  middle  point, 

EXERCISES. 

1.  Prove  that  the  bisectors  of  two  vertical  angles  are  in 
the  same  straight  line. 

2.  Prove  that  the  biseoiors  of  two  supplementary  adjacent 
angles  are  perpendicular  to  each  other. 

Note.— In  the  adjoining  figure  BOC  is  a  given  £^ 

angle;  and  one  of  its  arms  BO  is  produced  to  A;  N 

the  adjacent  supplementary  angles,  AOC,  BOC, 
are  bisected  by  EO,  DO. 

DO  and  EO  are  called  respectively  the  internal       . 

and  external  bisectors  of  the  angle  BOC.  O  " 

Hence  Exercise  2  may  be  stated  thus: 

Prove  that  the  internal  and  external  bisectoi's  of  an  angle  are  at  right  angles 
to  each  other. 


(GO) 


(Ax.  10) 
(CO) 

Q.E.D. 


BOOK  I.-PAUALLBL  LINES.  23 

3.  Show  that  the  angles  AOE  and  BOD  are  complemen- 
tary. 

4.  Show  that  the  angles  AOD  and  COD  are  supplemen- 
tary; and  also  that  the  angles  BOE  and  COE  are  supple- 
mentary. 

5.  If  the  angle  BOD  is  37°,  how  many  degrees  are  there 
in  AOE  ? 

6.  If  two  angles  are  supplementary,  and  the  greater  is  9 
times  the  less,  how  many  degrees  are  there  in  eacli  angle  ? 

7.  If  an  angle  is  11  times  its  complement,  how  many  de- 
grees does  it  contain  ?  ,  -^Z" 

Parallel  Lines. 

68.  Parallel  straight  lines  are  such  as  lie  in  the  same 
plane,  and  never  meet,  however  far  they  are  produced  in 
both  directions. 

69.  A  straight  line  crossing  sev- 
eral other  lines  is  called  a  transver- 
sal ;  as  EF. 

When  two  straight  lines  are  cut  by 
a  transversal,  eight  angles  are  formed, 
which  are  named  as  follows  : 

The  four  angles  a,  b,  (/,  h,  without 
the  two  lines,  are  called  exterior 
angles. 

The  four  angles  c,  d,  e,f,  within  the  two  lines,  are  called 
interior  angles. 

The  pair  c  and  e,  and  the  pair  d  and  /,  are  called  alter- 
nate-interior angles. 

The  pair  a  and  g,  and  the  pair  h  and  //,  are  called  alter- 
nate-exterior ^angles. 

The  pairs  a  and  e,  h  and  /,  c  and  g,  d  and  h,  are  called 
corresjjonding  angles.'^ 


*  Called  also  exterior-inierior  angles. 


24  PLANE  GEOMETRY. 


Proposition  lO.    Theorem. 

70.  Two  straight  lines  in  the  same  plane,  and  perpen- 
dicular to  a  third  straight  line,  are  parallel  to  each  other. 


Hyp.     Let  AB  and  CD  be  1  to  AC. 
To  prove  AB  and  CD  parallel. 

Proof.  If  AB  and  CD  are  not  parallel,  they  will  meet 
if  sufficiently  produced,  and  we  shall  have  two  perpendicu- 
lars from  the  same  point  to  the  same  straight  line,  which 
is  impossible.  (58) 

Therefore  they  cannot  meet. 

.  • .  AB  and  CD  are  parallel.  {G^) 

Q.E.I). 
EXERCISES. 

1.  Find  the  number  of  degrees  in  each  of  two  angles 
if  they  are  complementary,  and  the  greater  is  three  times 
the  less. 

2.  Find  the  number  of  degrees  in  each  of  two  angles  if 
they  are  supplementary,  and  the  greater  exceeds  the  less 
by  40°. 

3.  Find  the  number  of  degrees  in  each  of  two  angles  if 
they  are  supplementary,  and  the  less  is  one-fifth  the  greater. 


BOOK  L— PARALLEL  LINES.  25 


Proposition  1  1 .    Tlieorem. 

71.  //  a  straiglit  line  is  perpendicular  to  one  of  two  par- 
allels, it  is  also  ]jerpendicular  to  tUe  other. 


B 

-  E 


Hy2),  Let  AB  and  CD  be  two  parallel  lines,  and  let  AC 
be  1  to  CD. 

To  prove  AC  1  to  AB. 

Proof.     Through  A  where  AC  intersects  AB,  draw  AE 
1  to  AC 

Because  AE  and  CD  are  both  1  to  AC, 

.  • .  AE  is  parallel  to  CD.  (70) 

But  AB  is  parallel  to  CD.  (Hyp.) 

.  • .  AE  coincides  with  AB. 
TlirougU  a  gimn  pt.  only  one  line  can  he  drawn  \\ioa  given  line  (Ax.  12). 

•  •.  AB  is  JL  to  AC,  and  .-.  AC  is  J_  to  AB.         q.e.d. 

EXERCISES. 

1.  Find  the  value  of  an  angle  if  it  is  four  times  its  com- 
plement. 

2.  Find  the  value  of  an  angle  if  it  is  three  times  its  sup- 
plement. 

3.  Find  the  value  of  an  angle  if  it  is  one-eighth  of  its  com- 
plement. 

4.  Find  the  number  of  degrees  in  each  of  two  angles 
if  they  are  supplementary,  and  the  greater  is  four  times  the 
less.  X 


26  PLANE  OEOMETRY. 


Proposition  12.    Theorem. 

72.  If  a  straight  line  cut  tico  parallel  straight  lines,  the 
allernate-ititerior  angles  are  equal. 


Hyp,  Let  the  straiglit  line  LM  cut  the  ||  straight  lines 
AB,  CD,  at  the  points  E,  F. 

To  prove  /_  AEF  =  Z  EFD. 

Proof.  Through  K,  the  middle  point  of  EF,  draw 
HG  _L  to  AB. 

HG  is  also  _L  to  CD.  (71) 

Turn  the  figure  KEGD,  in  its  own  plane,  about  K  as  a 
pivot  until  KG  coincides  with  its  equal  KH. 

Then  since        /  FKG  =  Z  EKH,  being  vertical  Z  s,  (55) 

and  KF  =  KE,  (Cons.) 

.  • .  KF  will  coincide  with  KE, 
and  point  F  will  coincide  with  point  E. 
.  • .  GF  J_  to  KG  will  coincide  with  HE  _L  to  KH.     (58) 

.  • .   Z  s  KFG  and  KEH  coincide; 

.-.  Z  KFC=ZKEH.  Q.E.D. 

73.  Cor.  1.  The  alternate-interior  angles  BEF,  EEC 
are  also  equal,  (50) 

74.  Cor.  2.  The  alternate-exterior  angles  BEL,  CFM 
are  also  equal,  (55)  and  (Ax.  1) 


BOOK  I.-PARALLEL  LINES.  27 


Proposition  13.    Theorem. 

75.  Conversely,  if  a  straight  line  cut  tivo  other  straight 
lines,  so  as  to  make  the  alternate-interior  angles  equal,  the 
two  straight  lines  are  parallel. 


Hyp.  Let  the   straight    line   EF   cut  the   two  straight 
lines  AB,  CD  at  G  and  H,  so  that  ZAGH  =  ZGHD. 
To  prove  AB  parallel  to  CD. 

Proof.  Through  H  draw  KL  ||  to  AB. 
Then,    since   AB  and  KL  are  || , 

ZAGH=ZGHL.  (72) 

But  Z  AGH  =  Z  GHD.  (Hyp.) 

.-.  ZGHL  =  ZGIID.  (Ax.  1) 

.  • .  KL  and  CD  coincide. 

.-.CD  is  II  to  AB.  Q.E.D. 

76.  Cor.  If  the  alteimate-exterior  angles  are  eqiial,  the 
two  lines  are  parallel. 

If  the  sum  of  the  two  interior  angles  on  the  same  side  is 
less  than  two  right  angles,  the  lines  toill  meet. 

^  EXERCISE. 

If  a  line  is  drawn  through  the  vertex  of  an  angle  per- 
pendicular to  the  bisector,  prove  (1)  that  it  bisects  the 
supplementaiy  angle,  and  (2)  that  it  makes  equal  angles 
with  the  sides  of  the  given  angle. 


28 


PLANE  GEOMETRY. 


Proposition  14.    Theorem. 

77.  If  a  straight  line  cut  two  parallel  straight  lines,  (1) 
any  exterior  angle  is  equal  to  the  interior  oj^posite  angle,^ 
and  (2)  the  sum  of  the  two  ititerior  angles  on  the  same  side 
is  equal  to  two  right  angles. 

Hyp.  Let  the  st.  line  EF  cut  the  p 

two  II   st.  lines  AB,  CD  at  G  and  H.  

(1)  To  prove    Z  EGB  =:ZGHD.  ^ 
Proof.  Since    Z  AHG  =  Z  GHD,  — - 

being  alt. -int.  Zs  {12),  / 

and  ZAGH  =  ZEGB,  ^ 

being  vertical  /is  {Ti^), 
.'.  ZEGB  =ZGHD.  (Ax.  1) 

(2)  To  prove    Z  BGH  +  Z  GHD  =  2  rt.  Z  s. 

Proof  Z  EGB  =  Z  GHD.  (Just  proved. ) 

Add  Z  BGH  to  each. 

.-.  ZEGB  +  zBGH=rZGHD+  zBGH.     (Ax.  2) 
ZEGB+zBGH  =  2rt.  Zs, 
being  sup. -adj.  Zs  (52). 
.-.  Z GHD  +  Z  BGH  :=  2  rt. Z s.  (Ax.  1) 

Q.E.D. 


But 


EXERCISES. 


1.  If  EGB  =  42°,  how  many  degrees  are  there  in  CHF? 
inFHD?  in  BGH? 

2.  If  two   angles  are   supplementary,  and    the  greater 
exceeds  the  less  by  35°,  how  many  degrees  are  there  in  each 


angle  ? 


*  Calledcorresponding,  or  exterior-interior,  angles.    (09.) 


BOOK  I. -PARALLEL  LINES.  29 


Proposition  1  5.    Theorem. 

78.  Conversely,  if  a  straiglit  line  cut  two  other  straight 
lines,  (1)  so  as  to  make  the  exterior-interior  angles  equal, 
or  (2)  so  as  to  make  the  two  interior  angles  on  the  same  side 
together  equal  to  two  right  angles,  then  the  two  straight 
lines  are  ixirallel. 

(1)  Hyp.  Let  the  st.  line  EF  cut     

the  two  st.  lines  AB,  CD  at  G  and    ^ 

H,  so  that  Z  EGB  ==  Z  GHD.        H 


To  'prove  AB  ||  to  CD.  / 

Proof.  Since  Z  EGB  =  Z  GHD,  (Hyp.) 

and  ZEGB=ZAGH, 

heing  vertical  Z«  (55), 

.-.  ZAGH^ZGHD.  (Ax.  1) 

But  these  are  alternate  Z  s. 

.-.  AB  is  II  to  CD.  (75) 

(2)  Hyp.  Let  Z  BGH  +  Z  GHD  =  2  rt.  Z  s. 
To  prove  AB  ||  to  CD. 

Proof.  Since  Z  BGH  +  Z  GHD  =  2  rt.  Z  s,  ,    (Hyp.) 

and  ZBGH  +ZAGH  =  2  rt.  Zs, 

being  sup. -adj.  Zs  (53), 

.-.  ZBGH4-ZGHD  =  ZBGH  +  ZAGH.      (Ax.  1) 
Take  away  the  common  Z  BGH. 

.-.  ZGHD^ZAGH.  (Ax.  3) 

And  these  being  alternate  Z  s, 

,-.AB||toOa  (75) 

Q.E.D. 


A 

E/ 

B 

C 

Gi 

D 

H.l' 

30  PLANE  GEOMETRY. 


Proposition  1 6.    Theorem. 

79.  Straight  lines  winch  are  parallel  to  the  same  straight 
line  are  jJaraUel  to  one  a^iother. 

Hyp.     Let  the  st.  lines  AB,  CD 
be  each  ||  to  the  st.  line  PQ. 

To  prove  AB  ||  to  CD. 

Proof.     Draw  any  st.  line  EGH,       p 7*^*1 5 

cutting  the  lines  in  E,  G,  and  H.  / 

Since  AB  is  ||  to  PQ,  (Hyp.) 

.-.  Z  AEH  =  Z  EHQ, 

being  alt,  /s  (72)^ 

And  since  (ID  is  ||  to  PQ,  _(Hyp.) 

.-.  ZEGD  =  ZGHQ, 

being  ext.  -int.  Z  «  (77). 

.  • .  Z  AEHl  =  Z  EGD,  (Ax.  1) 

and  these  are  alternate  Z  s; 

.-.  ABislf^toCD.  (75) 

Q.E.D. 

Note.— If  PQ  lies  between  AD  and  CD,  the  Proposition  may  be  proved  in  a 
similar  way,  though  in  this  case  it  scarcely  needs  proof;  for  it  is  inconceivable 
that  two  straight  Unas,  which  do  not  meet  an  intermediate  straight  line,  should 
meet  each  other. 

EXERCISES. 

1.  Two  straight  lines  AB,  CD,  bisect  each  other  at  0. 
Show  that  the  straight  lines  joining  AC  and  BD  are  parallel. 

2.  If  a  straight  line  meets  two  parallel  straight  lines,  and 
the  two  interior  angles  on  the  same  side  are  bisected;  show 
that  the  bisectors  meet  at  right  angles.  * 


BOOK  L— PARALLEL  LINES.  31 


Proposition  17.    Theorem. 

80.  Two  angles  having  their  sides  parallel  each  to  each, 
are  either  equal  or  supplementary,  ^p, 

Hijp.     Let  AB  be  II  to  DQ,  and 
BC  be  II  to  MF. 

To  prove  Z  ABC  =  Z  DEF,  and     B  \ 
supplementary  to  Z  DEM.  .,    _ 

Proof,     Produce  BC  and  DH,  if        ^^\^ 
necessary,  until  they  intersect  in  Gr. 

Then  since  the  ||  s  AB  and  DE  are  cut  by  BC, 

.-.  ZABC=ZDGC.  (77) 

And  since  the  |1  s  BC  and  MF  are  cut  by  DH, 

.-.  ZDGC=ZDEF.  (77) 

.-.  ZABC  =  ZDEF.  (Ax.  1) 

Z  DEF  is  the  supplement  of  Z  DEM.  (25) 

.• .  Z  ABC  is  supplementary  to  Z  DEM.       q.e.d. 

81.  Cor.  Since  Z  DEF  =  Z  MEH,  being  vertical  Z  s,  (55) 
.*.  Z  ABC  =  Z  MEH,  and  is  supple7nentary  to  ZHEF. 

82.  Sen.  Two  parallels  are  said  to  be  in  the  same  direc- 
tion, or  in  op2JOsite  directions,  according  as  they  lie  on  the 
same  side  or  on  opposite  sides  of  the  straight  line  joining 
the  vertices.  Thus  AB  and  ED,  and  also  BC  and  EF, 
are  in  the  same  direction  because  they  lie  on  the  same  side 
of  BE.  But  BA  and  EH,  and  also  BC  and  EM,  are  in 
opposite  directions. 

Hence,  when  both  pairs  of  parallel  sides  are  either  in  the 
same  direction,  or  in  opposite  directions,  the  angles  are 
eqical.  But  when  two  of  the  parallel  sides  are  in  the  same 
direction,  and  the  other  two  in  opposite  directions,  the 
angles  are  supplementary. 


\ 


32  PLANE  GEOMETRY. 


Proposition  1 8.    Theorem. 

83.  Two  angles  liaving  their  sides  perpendicular  each  to 
each,  are  either  equal  or  supplementary. 


D 


Hyp.  Let  AB  be  1  to  GF,  and  AC  i.  to  DE. 
To  prove  /_  BAG  =  Z  DEF,  and  supplementary  to  Z  DEG. 
Proof.     Draw  AH  _L  to  AC,  and  AK  _L  to  AB. 
Then        AH  is  ||  to  ED,  and  AK  is  ||  to  EF. 
Two  lines  i  to  the  same  st.  line  are  \\  (70). 

Because  AH  and  AK  are  respectively  ||.to  ED  and  EF, 
and  extend  in  the  same  direction, 

.-.  ZHAK=:ZDEF.  (80) 

But         Z  HAK  is  the  complement  of  Z  HAB;      (Cons.) 

and         ZBAC  is  the  complement  of  ZHAB;      (Cons.) 

.-.ZHAK^ZBAC. 

Complements  of  equal  Z  s  are  equal  (50). 

.-.ZBAC   =ZDEF.  (Ax.  1) 

Since      Z  DEF  is  the  supplement  of  Z  DEG, 

.  • .  Z  BAC  is  supplementary  to  Z  DEG.  q.e.d. 

84.  ScH.  If  both  angles  are  acute  or  both  obtuse,  they 
are  equal ;  but  if  one  is  acute  and  the  other  obtuse,  they 
are  supplementary.  m 


BOOK  I.-TRlANOLEa. 


33 


TRIANGLES. 


85.  A  triangle  is  a  plane  figure  bounded  by  three 
straight  lines. 

The  three  straight  lines  which  bound  a  triangle  are  called 
its  sides.     Thus,  AB,  BC,  CA,  are  the 
sides  of  the  triangle  ABC. 

The  angles  of  the  triangle  are  the 
angles  formed  by  the  sides  with  each 
other;  as  BAG,  ABC,  ACB.  The  ver- 
tices of  these  angles  are  also  called  the 
vertices  of  the  triangle. 

86.  An  exterior  angle  of  a  triangle  is  the  angle  formed 
between  any  side  and  the  continuation  of  another  side; 
as  CAD. 

The  angles  BAC,  ABC,  BCA  are  called  interior  angles  of 
the  triangle.      When  we  speak   of  the 
angles  of  a  triangle  we  mean  the  three 
interior  angles. 

87.  An  equilateral  triangle  is  one 
whose  three  sides  are  equal. 

88.  An  isosceles  triangle  is  one  which 
has  two  equal  sides. 

89.  A  scalene  triangle  is  one  which 
has  three  unequal  sides. 

90.  A  right-angled  triangle  is  one 
which  has  a  right  angle.  * 

The  side  opposite  the  right  angle  is 
called  the  hypotenuse,  and  the  other 
two  sides  the  legs. 

91.  An  acute-angled  triangle  is  one 
which  has  tJiree  acute  angles. 

It  will  be  shown  hereafter  that  every  triangle  mu^t  have  at  least  two  acute 
angles, 


34 


PLANE  GEOMETRY. 


92.  An  obtuse-angled  triangle  is  one 
which  has  an  obtuse  angle. 

When  all  the  angles  are  equal  it  is 
called  an  equiangular  triangle. 

93.  The  base  of  a  triangle  is  the  side  upon  which  it  is 
supposed  to  stand. 

Either  side  may  be  taken  as  the  base;  but  in  an  isosceles 
triangle  the  side  which  is  not  one  of  the  equal  sides  is  called 
the  base. 

94.  When  one  side  of  a  triangle  has  been  taken  as  the 
base,  the  angle  opposite  is  called  the 
vertical  angle,  and  its  vertex  is  called 
the  vertex  of  the  triangle. 

95.  The  altitude  of  a  triangle  is  the 
perpendicular  let  fall  from  the  vertex 
to  the  base,  or  the  base  produced. 

Thus,  in  the  triangle  ABC,  AB  is  the  base,  ACB  is  the 
vertical  angle,  C  is  the  vertex,  and  CD  is  the  altitude. 

When  two  triangles  have  the  angles  of  the  one  equal  re- 
spectively to  the  angles  of  the  other,  the  angles  which  are 
equal  are  called  liomologons  angles,  and  the  sides  which 
are  opposite  the  equal  angles  are  called  homologous  sides ; 
hence  in  equal  triangles  the  homologous  sides  are  equal. 


Proposition  19.    Theorem. 

96.  Either  side  of  a  triangle  is  less  than  the  sum,  and 
greater  than  the  difference,  of  the  other  two  sides. 
In  the  A  ABC, 

A5  <  AC  +  BC,..      (Ax.  10) 
Taking  AC  from  each, 

AB  -  AC  <  BC, 
or  BO  >  AB  -  AC.  q.e.d. 


BOOK  L—TBIAN0LE8. 


Proposition  20.    Theorem. 

97.  The  sum  of  the  three  angles  of  any  triangle  is  equal 
to  two  right  angles. 

Hyp,  Let  ABO  be  any  A.  C  ,,£ 

To  prove 

ZA  +  zABC+ZC  =  2rt.Zs. 

Proof.     Produce     AB    to    T), 
and  draw  BE  |)  to  AC. 

Then  since  BE  is  I|  to  AC, 

.  • .  ext.  Z  EBD  =  int.  Z  A,  (77) 

and  ZCBE  =  alt.  ZC  (72) 

Add  ZABCtoeach. 

.  • .  Z  ABC+  Z  CBE+  Z  EBD  =  Z  ABC+  Z  C+  Z  A.  (Ax.  2) 

But  ZABC  +  ZCBE  +  ZEBD  =  2rt.  Zs. 

The  sum  of  all  the  /_son  the  same  side  of  a  st.  line  at  a  point  =  2rt.  Zs  (53). 

.-.  ZABC+ZC+ZA  =  2rt.  Zs.  (Ax.  1) 

Q.E.D. 

98.  CoE.  1.  Since  Z  EBD  =  Z  A,  and  Z  CBE  =  Z  C, 

.•.ZCBD=ZA-hZC. 
Hence^  the  exterior  angle  of  a  triangle  is  equal  to  the  sum 
of  the  two  opjjosite  interior  angles. 

99.  CoE.  2.  If  two  a7igles  of  a  triangle  are  given,  or 
merely  their  sum,  the  third  angle  can  he  found  hy  subtract- 
ing  this  sum  from  two  right  angles. 

100.  CoE.  3.  If  two  triangles  have  iivo  angles  of  the  one 
equal  to  two  angles  of  the  other,  the  third  angles  are  equal. 

101.  CoE.  4.  A  triangle  can  have  hut  one  right  angle, 
or  hit  one  obtuse  angle. 

102.  Coe".  5.  In  any  right-angled  triangle  the  two  acute 
angles  are  complementary. 

103.  CoE.  6.  Each  angle  of  an  equiangular  triarigle  t> 
two-thirds  of  a  right  angle. 


36 


PLANE  GEOMETRY. 


Proposition  21.    TFieorem. 

104.  Two  triangles  are  equal  when  two  sides  and  the 
included  angle  of  the  one  are  equal  respectively  to  two  sides 
and  the  included  angle  of  the  other. 

Hyp,    Let  ABC,  DEF   be  two 
AS,  having  AB  =  DE,  AC  =  DF. 
ZA  =  ZD. 

To  prove   A  ABC  =  A  DEF. 

Proof,  Apply  the  A  ABC  to  the 
A  DEF  (29)  so  that  the  point  A 
shall  fall  on  D,  and  the  side  AB  on  DE. 

Then  because       AB  =  DE  (Hyp.), 

.  • .  pt.  B  must  fall  on  E. 
Because  Z  A  =  Z  D  (Hyp.). 

.  • .  AC  must  lie  along  DF. 
And  because         AC  =  DF  (Hyp.), 

.  * .  pt.  C  must  fall  on  F. 
Now,  since  B  falls  on  E,  and  C  on  F, 

.  •.  BC  must  coincide  with  EF.     (Ax.  11) 
Hence,  since  BC  coincides  with  EF, 

.  • .  BC  =  EF,  Z  B  =  Z  E,  Z  C  =  Z  F. 

Therefore  the  two  As  coincide  in  all  their  parts,  and 
hence  are  equal  (29).  '         Q.E.D. 


Note.— Wlien  all  the  parts  of  one  triangle  are  respectively  equal  to  all  the 
parts  of  another  triangle,  the  triangles  are  said  to  be  "equal  in  all  respects." 
Such  triangles  are  said  to  be  identically  equal,  or  congruent. 

This  proposition  is  proved  by  "  the  method  of  superposition,"  i.e.,  it  is  shown 
that  one  of  the  triangles  could  be  placed  on  the  other  so  as  to  cover  it  exactly 
without  overlapping;  and  then,  since  all  their  parts  would  coincide,  the  two  tri- 
angles are  equal  by  definition  (29). 


BOOK  I.— TRIANGLES. 


37 


Proposition  22.    Theorem. 

105.  Ttvo  triangles  are  equal  when  a  side  and  tlie  two 
1 1  adjacent  angles  of  the  one  are  eqnal  respectively  to  a  side 
11  and  the  two  adjacent  angles  of  the  other, 
til       Hyp,  Let  ABC,  DEF  be  two 
'      AS  having  ZA=:ZD,  ZB  = 
Z  E,  AB  =  DE. 
To  prove    A  ABC  =  A  DEF. 
Proof.  Apply  the   A  ABC  to 
the  A  DEF  so  that  the  point  A 
shall  fall  on  D,  and  the  side  AB 
onDE. 


Then, 


(Hyp.) 


Because 


Because 


(Hyp.) 


(Hyp.) 


because  AB  =  DE, 
.  • .  pt.  B  must  fall  on  E. 

Z  A  =  Z  D, 
.  • .  AC  must  lie  along  DF. 

Z  B  =  Z  E, 

.•.  BC  must  lie  along  EF. 

Because  AC  and  BC  fall  upon  DF  and  EF,  respectively, 
.  • .  the  point  C,  falling  upon  both  lines  DF  and  EF,  must 
fall  at  their  point  of  intersection  F. 

.  • .  the  two  A  s  coincide  in  all  their  parts,  and  are  equal. 

Q.E.D. 

106.  Cor.  1.  Tiuo  right-angled  triangles  are  equal  luhen 
the  hypotemise  a7id.  a7i  acute  angle  of  the  one  are  equal 
respectively  to  the  hypotenuse  and  an  acute  angle  of  the 
other, 

107.  Cor.  2.  Two  right-angled  triangles  are  equal  when 
a  side  and  an  acute  angle  of  the  one  are  equal  respectively 
to  a  side  ayid  homologous  acute  a7igle  of  the  other. 


38 


PLANE  GEOMETRY, 


Proposition  23.    Theorem. 

108.  Two  triangles  are  equal  if  the  three  sides  of  the 
me  are  equal  respectively  to  the  three  sides  of  the  other, 

Hijjp,  Let  ABC,  DEF  be 
two  AS  having  AB  =  DE, 
AC  =  DE,  BC  =  EF. 

To  prove  A  ABC  =  A  DEE. 

Proof  Apply  the  A  ABC 
to  the  A  DEE  so  that  the  side 
AB  shall  coincide  with  its  equal 
DE,  and  the  vertex  C  fall  at 
F'  on  the  opposite  side  of  DE 
from  F. 

Join  EF'  which  cuts  DE  at  H. 

Because        DF  =  DE',     and     EF  =  EF',  (Hyp.) 

.*.  points  D  and  E  are  equally  distant  from  F  and  F'. 

.-.  DE  is  _L  to  EF'  at  its  middle  pt.  H.  (67) 

Now  revolve  the  A  DEF  about  DE  as  an  axis  till  it 
comes  into  the  plane  of  the  A  DEE'. 

Because       /  DHF  =  /_  DUE'  =  a  rt.  Z  (67) 

and  HF  =  HE', 

.-.  pt.  F  will  fall  on  F', 

.  • .  the  two  A  s  coincide  in  all  their  parts,  and  are  equal. 

Q.E.D. 

109.  ScH.  When  two  triangles  are  equal,  the  equal 
angles  lie  opposite  the  equal  sides;  and  conversely,  the 
equal  sides  lie  opposite  the  equal  angles.  Thus,  the  angles 
A  and  D  are  equal  or  homologous  angles  (95).  Also  the 
angles  C  and  F  are  homologous;  likewise  the  angles  I^ 
and  E. 




^Krhypotenuse  and  a  side  of  the  one  are  equal  respectively  to 
the  hypotenuse  and  a  side  of  the  other. 


BOOK  L— TRIANGLES.  39 


Proposition  24.    Theorem. 


I 


Hyp,  Let  ABC,  DEF  be  two  rt.  A  s  having 
hyp.  AC  =  DF,  and  BC  =  EF. 
To  prove       A  ABC  =  A  DEF. 

Proof  Apply  the  A  ABC  to  the  A  DEF  so  that  BC 
shall  coincide  with  its  equal  EF. 

Then,  since  Z  B  =  Z  E  =  a  rt.  Z , 

.*.  BA  will  lie  along  ED,  and  pt,  A  will  fall  somewhere 
on  ED. 

But  the  equal  oblique  lines  CA  and  FD  cut  off  on  ED 
equal  distances  from  the  foot  of  the  J_  EF  (61). 

.-.  pt.  A  will  fall  at  D. 

.-.  the  two  A  s  coincide  in  all  their  parts,  and  are  equal. 

Q.E.D. 


EXERCISES, 


1.  If  BC,  the  base  of  an  isosceles  triangle,  ABC,  is  pro- 
duced to  any  point  D,  show  that  AD  is  greater  than  either 
of  the  equal  sides. 

2.  Prove  that  the  sum  of  the  distances  of  any  point  from 
the  three  vertices  of  a  triangle  is  greater  than  half  its 
perimeter. 

3.  If  one  of  the  acute  angles  of  a  right  triangle  is 
40°  14'  48",  what  is  the  value  of  the  other  acute  angle  ? 


40 


PLANE  GEOMETRY. 


Proposition  25.    Theorem. 

111.  In  an  isosceles  triangle  the  angles  opjjosite  the  equal 
sides  are  equal. 

Hyp.  Let  ABC  be  an  isosceles  A  having 
AC  =  BC. 

To  prove         Z  A  =  Z  B. 

Proof.  Draw  the  line  CD  from  the 
vertex  C,  to  the  middle  pt.  D  of  the 
base. 


Then,  in  the  As  ADC  and  BDC, 
because  AC  =  BC, 

AD  =  DB, 


(Hyp.) 
(Cons.) 


CD  =  CD,    (being  common  to  both  As,) 

.-.  A  ADC  =  A  BDC. 
Two  AS  are  equal  if  the  three  sides  are  equal  each  to  each  (108). 


{ 


.'.  Z  A  =  Z  B, 
being  opposite  equal  sides  (109). 


Q.E.D. 


112.  Cor.  1.  Since  A  ADC  =  A  BDC,  we  have  Z  ACD 
=  Z  BCD,  and  Z  ADC  =  Z  BDC.  Therefore,  the  straight 
line  which  joins  the  vertex  to  the  middle  jjoint  of  the  hase  of 
an  isosceles  triangle,  is  at  right  a?igles  to  the  base,  and  bisects 
the  vertical  angle. 

Hence,  also,  the  bisector  of  the  vertical  angle  of  an  isosceles 
triangle  bisects  the  base  at  right  angles. 

113.  Cor.  2.  TJie  j)erpendicular  from  the  vertex  to  the 
base  of  an  isosceles  triangle  bisects  the  base  and  the  angle  at 
the  vertex. 

An  equilateral  triangle  is  also  equiangular. 


BOOK  L-T1UANGLE8. 


41 


Proposition  26.    Theorem. 

114.  Conversely,  if  two  angles  of  a  triangle  are  equal, 
the  sides  opposite  the  equal  angles  are  equal,  and  the  triangle 
IS  isoceles. 

Htjp.   Let  ABC  be  a  A  having 

ZA=ZB. 

To  prove  AC  =  BC. 

Proof.  Draw  CD,  bisecting  the  Z  ACB. 

Then,  in  As  ACD,  BCD, 


Also, 


ZA=.  ZB. 

(Hyp.) 

ZACD:=  ZBCD. 

(Cons.) 

.-.  ZADC=  ZBDC. 

(100) 

CD  is  common. 

.-.  aACD=  a  BCD, 

having  a  side  and  the  two  adjacent  angles  equal,  each  to  eoAili  (105). 


.  • .  AC  ==  BC, 

Jyeing  homologous  sides  of  equal  as  (109). 


Q.E.D. 


115.  Coil.  Au  equiangular  triangle  is  also  equilateral. 


EXERCISES. 

1.  ABC  is  an  isosceles  triangle,  with  AB  =  AC.  The 
bisectors  of  the  angles  B  and  C  meet  at  0.  Prove  that 
CO  =  BO.      - 

2.  ABC  is  a  triangle ;  BA  is  produced  to  D  so  that 
AD  =  AC,  and  DC  is  joined.    Prove  that  Z  BCD  >  Z  BDO. 

3.  The  angle  C  is  twice  as  large  as  either  of  the  angles  A 
and  B  :  how  many  degrees  are  there  in  each  angle  ? 


42  PLANE  GEOMETRY. 


Proposition  27.    Theorem. 

116.  If  one  side  of  a  triangle  he  greater  than  another, 
the  angle  opposite  the  greater  side  is  greater  than  the  ayigle 
opposite  the  less, 

c 


Hyp,  Let  ABC  be  a  A  having  the  side 

AC>BC. 

To  prove  ZABC>ZA. 

Proof  From  CA  cut  off  CD  =  CB. 
Join  BD, 

Then,  since  CD  =  CB, 

.-.  ZCDB=:  zCBD.  (Ill) 

But  ext.  ZCDB  >  opposite  int.  /DAB  of  aADB.   (98) 

.-.  also  ZCBD>ZDAB. 

But  Z  CBA  >  Z  CBD.  (Ax.  8) 

Much  more  .*.  ZCBA>ZDAB.  q.e.d. 

EXERCISES. 

1.  Prove  Prop.  27  by  producing  CB  to  E,  making  CE 
=  CA. 

2.  ABO  is  a  triangle  in  which  OB,  00  bisect  the  angles 
B,  0,  respectively;  show  that,  if  AB  is  greater  than  AC;, 
then  OB  is  greater  than  00. 


BOOK  I.-TRIANOLES.  43 


Proposition  28.    Theorem. 

117.  Conversely,  if  one  angle  of  a  triangle  he  greater 
than  another,  the  side  opposite  the  greater  angle  is  greater 
than  the  side  opposite  the  less. 

Hyjy.  Let  ABC  be  a  A  having 

Z  ABO  >  Z  BAG. 
To  prove  AC>BC. 

Proof.  Draw  BD  so  that 

ZABD=  ZBAD. 

Then  AD  =  BD, 

being  opposite  equal  Zs  (114). 
« 

But  in  A  BCD,  BD  +  DC  >  BC. 

Mtlier  side  of  a  A  is  <  the  sum  of  the  oilier  tioo  (96). 

And,  since  AD  =  BD, 

.  • .  BD  +  DC  =  AC. 

.-.  AC>BC.  Q.E.D. 

Note. — These  two  propositions  may  be  stated  briefly  as  follows:  In  every 
triangle,  the  greater  side  is  opposite  the  greater  angle,  and  conversely,  the 
gi  eater  angle  is  opposite  the  greater  side. 

118.  Cor.  The  hypotenuse  is  the  greatest  side  of  a 
right-angled  triangle. 


EXERCISES. 

1.  In  a  A  ABC,  it  AC  is  not  greater  than  AB,  show  that 
any  straight  l^e  drawn  through  the  vertex  A  and  termi- 
nated by  the  base  BC,  is  less  than  AB. 

2.  Any  two  sides  of  a  triangle  are  together  greater  than 
twice  the  straight  line  drawn  from  the  vertex  to  the  middle 
point  of  the  third  side. 


44 


PLANE  GEOMETRY. 


Proposition  29.    Theorem. 

119.  i/*  two  triangles  have  two  sides  of  the  one  equal 
respectively  to  two  sides  of  the  other,  but  the  iiicluded  angle 
171  the  first  triangle  greater  than  the  included  angle  in  the 
second,  then  the  third  side  of  the  first  triangle  is  greater 
than  the  third  side  of  the  second. 


Hyp,  Let  ABC,  DEF  be  two  A  s,  having 
AB  =  DE,    AC  =  DF, 

but  ZBAC>ZEDF. 

To  prove  BC>EF. 

Proof.  Apply  the  A  ABC  to  the  A  DEF  so  that  AB  shall 
coincide  with  DE,  and  the  pt.  C  fall  at  H. 

Join  FH. 

Then,  .  since  DH  =  DF, 

.-.  ZDHF=  ZDFH. 

But  ZDHF>ZEHF. 
The  wlwU  18  greater  than  any  of  its  parts  (Ax.  8). 

.-.  also  zr)FH>zEHF. 
But  ZEFH>zDFH. 
Much  more  .  • .  Z  EFH  >  Z  EHF. 

.-.  EH>EF. 
TTie  greater  side  is  opposite  the  greater  angle  (117). 
But  EH  =  BC.    .  • .  BC  >  EF.  q.e.d. 


(Hyp.) 
(Ill) 


(Ax.  8) 


BOOK  L— TRIANGLES,  45 


Proposition  30.    Theorem. 

120.  Conversely,  if  tivo  triangles  have  two  sides  of  the 
one  equal  respectively  to  two  sides  of  the  other,  hut  the  third 
side  of  the  first  triangle  greater  than  the  third  side  of  the 
second,  then  the  included  angle  of  the  first  triangle  is  greater 
than  the  included  angle  of  the  secotid. 

Hyp.     Let  ABC,  DEF  be  two  As 
having  AB  =  DE,  AC  =  DF, 

but  BC  >  EF. 

To  prove       Z  A  >  Z  D. 
Proof       If  Z  A  is  not  >  iT), 

then  either  (1)    Z  A  =  ZD, 

or  (2)    ZA<  ZD. 

(1)  If  ZA=  zD, 
then  BC  =  EF, 

Immng  two  sides  and  the  included  Z  equal,  each  to  each  (104). 
But  this  is  contrary  to  the  hypothesis. 

(2)  If  ZA<ZD, 

then  BC<EF.  (119) 

But  this  is  contrary  to  the  hypothesis. 
Hence,  since  Z  A  can  neither  be  —  nor  <  /_T), 

.-.    ZA>  ZD.  Q.E.D. 

Note.— Prop.  30  is  here  proved  by  an  indirect  method.  This  method  is 
sometimes  called  the  reductio  ad  absurdum.  It  consists  in  assuming  that  the 
conchision  to  be  proved  is  not  tnie,  and  showing  tliat  this  assumption  leads  to 
an  absurdity,  or  to  a  result  inconsistent  with  the  hypothesis.  / 


46  PLANE  GEOMETRY. 


Quadrilaterals. 

121.  A  quadrilateral  is  a  plane  figure  bounded  by  four 
straight  lines,  which  are  called  its  sides. 

The  straight  lines  which  join  opposite  angles  of  a  quadri- 
lateral are  called  the  diagonah. 

122.  A  trapezium  is  a  quadrilateral 
which  has  no  two  of  its  sides  parallel. 

123.  A  trapezoid  is  a  quadrilateral 
which  has  two  of  its  sides  parallel. 

The  parallel  sides  of  a  trapezoid  are 
called  the  bases,  and  the  perpendicular 
distance  between  them  is  called  the 
altitude.  The  line  joining  the  middle 
points  of  the  non-parallel  sides  is  called 
the  middle  parallel*'  of  the  trapezoid. 

124.  A  parallelogram  is  a  quadri- 
lateral which  has  its  opposite  sides 
parallel.  . 

The  bases  of  a  pafallelogram  are  the 
side  on  which  it^ands  and  the  opposite 
side.  The  perpendicular  distance  be- 
tween the  bases  is  called  the  altitude. 

125.  A  rectajigle  is  a  parallelogram 
whose  angles' are  right  angleXf^ 

126.  A  square  is  a  rectangle  whose 
sides  are  all  equal. J 

127.  A  rhomboid  is  a  parallelogram 


whose  angles  are  oblique  and  whose  ad- 
jacent sides  are  unequal. 

128.  A  rhombus,  or  lozenge,  is  a  par- 
allelogram whose  sides  are  all  equal.  § 


*  Called  also  the  median. 
t  Called  also  a  right-angled  parallelogram. 
X  Callt^d  also  an  equilateral  rectangle. 
§  Called  also  equilateral  rhomboid. 


BOOK  I— QUADRILATERALS.  47 


Proposition  3 1 .    Theorem. 

129.  1)1  every  parallelogram,  the  op2)osite  sides  are  eqirnl, 
and  the  opposite  angles  are  eqicah 

Hyp.  LetABCDbea^.  D __^ 

To  prove  DC  =  AB,     AD  =  BO,     \  ^>  ^ 

Z.K^  IQ,    ZD=.ZB.       \o^^^ 


\ 


Proof.  Draw  the  diagonal  AC.                ^ 

B 

Because                         DC  is  ||  to  AB, 

(Hyp.) 

and                               AD  is  II  to  BO, 

(Hyp.) 

.-.  ZDCA=  ZBAC, 

and                    ^        ZACB=ZCAD, 

being  alt 4nt.  Is  (73). 

Hence  the  whole        Z  DCB  =  Z  BAD.  (Ax.  9) 

Now  in  the  As  ACD,  ACB,  because 

(ZDCA=  ZBAC,) 
(ZCAD=ZACB,[      (,l"st  proved) 

and  AC  is  common, 

.-.   A  ACD  =  A  ACB, 
having  a  side  and  the  two  adjacent  Z,s  equal,  each  to  each  (105) 

.-.  DC  =  AB,  AD  =BC, 
being  liomologous  sides  of  equal  as  (109). 

and  Z  D  =  Z  B.  q.e.  d. 

*> 

130.  Cor.  1.  A  diagonal  of  a  parallelogram  divides  it 
into  two  equal  triangles. 

131.  Cor.  2.    Two  parallels  included  between  tivo  other 
parallels  are  equal. 


48  PLANE  GEOMETRY. 


Proposition  32.    Theorem. 

132.  If  the  O'piiosite  sides  of  a  quadrilateral  are  equal, 
the  figure  is  a  parallelografu. 

Hyp.  Let  ABCD  be  a  quadrilat- 
eral having 

AB  =  CD,  AD  =  BC. 

To  prove  ABCD  is  a  CJ. 

Proof.  Draw  the  diagonal  AC. 
In  the  As  ACD,  ACB,  because 

AD  =  BC,  (Hyp.) 

AB  =  CD,  (Hyp.) 

AC  is  common, 

.-.    aACD=  a  ACB.  (108) 

.-.    ZACD  =  ZCAB, 

and  Z  CAD  =  Z  ACB, 

being  homologous  /s  of  equal  as  (109). 

.*.  ABisli  to  CD, 

and  AD  is  II  to  BC, 

tJie  alt. -int.   Zs  being  equal  (75). 

.  • .  ABCD  is  a  fU  by  definition.  (124) 

Q.E.D. 
EXERCISES. 

1.  If  one  angle  of  a  parallelogram  is  a  right  angle,  prove 
that  all  its  angles  are  right  angles. 

2.  Prove  that  two  parallels  are  everywhere  equally  distant. 

3.  If,  in  the  figure  of  Prop.  32,  BE  be  drawn  parallel  to 
AC  and  meeting  DA  produced  to  E,  prove  that  the  paral- 
lelogram EBCA  will  be  equal  to  the  parallelogram  ABCD. 


BOOK  L-QUADRILATERAL8.  49 


Proposition  33.    Theorem. 

133.  If  tivo  opposite  sides  of  a  qttadrilaterdl  are  equcd 
ayid  parallel,  the  figure  is  a  parallelogram. 

Hyp,  Let  ABCD  be  a  quadrilat-    D_ 
eral,  having  AB  =  and  ||  to  DC. 

To  prove  ABCD  is  a  m. 

Proof.     Draw  the  diagonal  AC. 


In  the  A  s  ACD,  ACB,  because 

AB  =  CD,  (Hyp.) 

AC  is  common, 

ZACD=  ZCAB, 

being  alt. -int.  /.  8  {12), 

.-.  aACD=  aACB, 
Jiamng  two  sides  and  the  included  A  equal,  each  to  each  (104). 

.  • .  AD  =  BC, 

being  homologous  sides  of  equal  as  (109). 

.  * .  the  figure  ABCD  is  a  CD, 
Jiamng  its  opposite  sides  equal  (132).  q.e.d. 

EXERCISES. 

1.  Prove  that  the  straight  lines  which  bisect  two  adjacent 
angles  of  a  parallelogram  cut  each  other  at  right  angles. 

2.  AB,  CD,  EF  are  three  equal  and  parallel  straight 
lines ;  prove  that  the  triangle  ACE  is  equal  to  the  triangle 
BDF. 


50  PLANE  GEOMETRY, 


Proposition  34.    Theorem. 

134.  The  diagonals  of  a  parallelogram  Used  each  other. 

Hyp.  Let  ABCD  be  a  /Z7,  whose 
diagonals  intersect  at  0. 

To  prove     AO  =  OC,  DO  =  OB. 

Proof.  lu  the  As  AOB,  COD,  "  ^ 

AB  =  DC, 

hdnig  opp.  sides  of  the  OJ  (129), 

ZABO=  ZCDO, 

and  ZBAO=  ZDCO, 

being  alt. -int.  As  (72). 

.-.    aAOB=  aCOD, 

having  a  side  and  the  two  adj.  Zs  equal,  each  to  each  (105). 

.  • .  AO  =  OC,  and  DO  =  OB.  (109) 

Q.E.I). 
EXERCISES. 

1.  If  the  opposite  angles  of  a  quadrilateral  are  equal,  the 
figure  is  a  parallelogram. 

2.  If  the  diagonals  of  a  quadrilateral  bisect  each  other, 
the  figure  is  a  parallelogram. 

3.  If  the  diagonals  of  a  parallelogram  are  equal,  the  fig- 
ure is  a  rectangle ;  if  they  also  intersect  at  right  angles,  it  is 
a  square. 

4.  The  straight  lines  joining  the  middle  points  of  the 
opposite  sides  of  any  quadrilateral  bisect  each  other. 

5.  The  diagonals  of  a  rhombus  bisect  each  other  at  right 
angles. 

G.  The  diagonals  of  a  rectangle  are  equal. 


BOOK  L^qUADRILATEIlAL8,  51 


Proposition  35.    Theorem. 

135.  Two  parallelograms  are  equal  when  two  adjacent 
sides  and  the  included  angle  of  the  one  are  equal  respectively 
to  two  adjacent  sides  and  the  included  angle  of  the  other. 

Hyp,  Let  ABCD,  A'B'C'D'  be  two 
^^s,  having  AB  =  A'B', 

AD  =  A'D',  V  \ 

ZA  =  ZA'.  \ \ 

To  prove  ru ABCD  =diA'B'G'D\   jy c' 

Proof            Apply   ZZ7ABCD    to  \                       \ 

^A'B'C'D',  so  that  Z  A  shall  coin-         \ \^ 

cide    with  Z  A',   and   the   side  AB  a'                       b' 
with  the  equal  side  A'B'. 


Because       Z  A  =  Z  A',  and  AD  =  A'D',  (Hyp. ) 

.-.  AD  will  fall  on  A'D',  and  pt.  D  on  pt.  D'. 

Because    DC  is  ||  to  AB,  and  D'C  is  ||  to  A'D',        (124) 

.-.  DC  will  fall  on  D'C,  and  pt.  C  somewhere  on  D'C. 
Through  a  given  pt.  only  one  st.  I.  can  be  drawn  \\  to  a  given  st.  I.  (Ax.  12) 

Also,  because  BC  is  ||  to  AD,  and  B'C  is  ||  to  A'D',  (124) 

.♦.  BC  will  fall  on  B'C,  and  pt.  C  somewhere  on  B'C.  (Ax.  12) 

.*.  since  pt.  C  falls  on  D'C  and  B'C,  it  must  fall  at  their 
pt.  of  intersection  C. 

.'.  the  two£I7s  coincide  throughout,  and  are  equal,  q.e.d. 

136.  COK.  Two  rectangles  are  equal  ivhen  they  have  two 
adjacent  sides  eqxial,  each  to  each. 


52 


PLANE  GEOMETRY, 


POLYGOl^S. 


137.  A  polygon  is  a  plane  figure  bounded  by  straight 
lines;  as  ABODE. 

The  straight  lines  are  called  the  sides 
of  the  polygon;  and  their  sum  is  called 
the  perimeter  of  the  polygon. 

The  angles  of  the  polygon  are  the 
angles  formed  by  the  adjacent  sides  with 
each  other  ;  and  the  vertices  of  these 
angles  are  also  called  the  vertices  of  the 
polygon. 

138.  The  angles  of  the  polygon  measured  on  the  side 
of  the  enclosed  surface  are  called  interior  angles. 

An  exterior  angle  of  a  polygon  is  an  angle  between  any 
side  and  the  continuation  of  an  adjacent  side. 

A  diagonal  is  a  line  joining  any  two  vertices  that  are  not 
consecutive;  as  AD. 

139.  Polygons  are  named  from  the  number  of  their 
sides,  as  follows : 

A  polygon  of  three  sides  is  a  triangle;  one  of  four  sides, 
a  quadrilateral;  one  of  five  sides,  a  pentagon;  one  of  six 
sides,  a  hexagon;  one  of  seven  sides,  a  heptagon;  one  of  eight 
sides,  an  octagon;  one  of  nine  sides,  a  nonagon;  one  of  ten 
sides,  a  decagon;  one  of  twelve  sides,  a  dodecagon;  one  of 
fifteen  sides,  a  quindecagon. 

140.  An  equilateral  polygon  is  one  which  has  all  its 
sides  equal.     An  equiangular  polygon 
is  one  which  has  all  its  angles  equal. 
A  re^i^Z^Mi- polygon   is   one  which  is 
both  equilateral  and  equiangular. 

141.  A  cow_ye^_polygon  is  one  each 
of  whose  interior  angles  is  less  than  a 
straight  angle;  as  ABODE. 


BOOK  L— POLYGONS.  t)6 

142.  A  concave  polygon  is  one  in  which  at  least  one  of 
the  interior  angles  is  reflex  (26),  as 
GHKMNO,    in    which    the   interior 
angle  KMN  is  reflex.* 

The    polygons   considered   in   this   g^ 
work  will  be  understood  to  be  convex, 
unless  otherwise  stated. 

It  is  evident  that  a  polygon  has  as  many  angles  as  sides. 

143.  Two  polygons  are  mutually  equilateral  when  the 
sides  of  the  one  are  equal  respectively  to  the  sides  of  the 
other,  taken  in  the  same  order;  as 
the  polygons  ABCD,  A'B'C'D',  in 
which  AB  =  A'B',  BC  =  B'C,  etc. 

144.  Two  polygons  are  mutually 
equiangular  when  the  angles  of  the 
one  are  equal  respectively  to  the 
angles  of  the  other,  taken  in  the 
same  order;  as  the  polygons  PQRS, 
P'Q'R'S',  in  which  Z  P  =  Z  P', 
ZQ  =  ZQ',  etc. 

145.  In  polygons  which  are  mutually  equilateral  or  mu- 
tually equiangular,  any  two  corresponding  sides  or  angles 
are  called  homologous  (95). 

Except  in  the  case  of  triangles,  two  polygons  may  be  mu- 
tually equilateral  without  being  mutually  equiangular,  and 
mutually  equiangular  without  being  mutually  equilateral. 

146.  A  polygon  may  be  divided  into  triangles  by  draw- 
ing diagonals  from  one  of  its  vertices  ;  and  the  number  of 
triangles  into  which  any  polygon  can  thus  be  divided  is 
evidently  equal  to  the  number  of  its  sides,  less  two. 

When  two  po^gons  can  be  divided  by  diagonals  into  the  same  number  of  tri- 
angles, equal  each  to  each,  and  similarly  placed,  the  polygons  are  equal ;  for 
they  can  be  applied  one  to  the  other,  and  the  corresponding  triangles  will  evi- 
dently coincide,  and  therefore  the  polygons  w  ill  coincide  throughout. 

When  two  polygons  are  both  mutually  equilateral  and  mutually  equiangular. 
they  are  equal,  for  they  can  be  applied  one  to  the  other  so  as  to  coincide. 

*  Called  also  a  reentrant  angle. 


54  PLANE  GEOMETRY, 


Proposition  36.    Theorem. 

147.  The  sum  of  the  interior  angles  of  a  polygon  is 
equal  to  two  right  angles  taken  as  many  times  less  two  as 
the  polygon  has  sides, 

F  E 


,/.•(*■''' 


^' — ->»  ]j^^-^'- 


Hyp,  Let  ABCDEF  be  a  polygon  having  n  sides. 
To  prove  ZA+zB+zC  +etc.  =  2  rt.  Zs  {n-2). 
Proof  From  any  vertex  A,  draw  the  diagonals  AC, 
AD,  AE.  The  polygon  will  be  divided  into  as  many 
triangles  less  two  as  it  has  sides  (146).  Thus,  there  are 
(n  —  2)  triangles,  whose  angles  make  the  interior  angles  of 
the  polygon. 

Now,  the  sum  of  the  Z  s  of  a  A  =  2  rt.  Z  s.  (97) 

.*.  the  sum  of  the  Z  s  in  the  polygon  =  2  rt.  Z  s  (/^  —  2). 

Q.E.D. 

148.  Cor.  1.  2  rt.  Zs  (/^  -  2)  -  2?^  rt.  Zs  -4  rt.  Zs. 
Therefore,  the  sum  of  the  angles  of  a  polygon  is  also  equal 

to  tivice  as  many  right  angles  as  the  figure  has  sides,  less 
four  right  angles. 

149.  Cor.  2.  Tlie  sum  of  the  angles  of  a  quadrilateral  is 
equal  to  two  right  angles  taken  (A  — 2)  times,  i.e.,  four 
right  angles.  TJie  sum  of  the  angles  of  a  pentagon  is  equal 
to  two  right  angles  talcen  (5  —  2)  times,  i.e.,  six  right 
a7igles,  etc. 

150.  Cor.  3.  Each  ayigle  of  an  equiangular  polygon  of 

fi,  sides  is    ^     — -  right  angles. 


BOOK  I.^POLYGONS.  65 


Proposition  37.    Theorem. 

151,  The  sum  of  the  exterior  angles  of  a  polygon,  made 
hy  producing  each  of  its  sides  in  the  smne  direction,  is  equal 
to  four  right  angles. 


Hyp.  Let  ABODE  be  a  polygon  having  its  n  sides  pro- 
duced in  the  same  direction. 

To  prove  Zci  +  Zl)  +  Z.c+  etc.  =  4  rt.  Z  s. 
Proof  Any  int.  Z  A  +  its  adj.  -ext.  Z  «^  =  3  rt.  Z  s,       (52) 
. • .  all  the  int. Z s  +  all  the  ext. Zs  =  2nrtZ s. 
But  all  the  int.  Z  s  =  27^  rt.  Z  s  -  4  rt.  Z  s.  (148) 

.  • .  all  the  ext.  Z  s  =  4  rt.  Z  s.  (Ax.  3) 

Q.E.D. 

EXERCISES. 

1.  Express  in  terms  of  a  right  angle,  and  also  in  degrees, 
the  magnitude  of  each  interior  angle  of  (1)  a  regular  hexa- 
gon, (2)  a  regular  octagon,  and  (3)  a  regular  decagon. 

2.  If  one  side  of  a  regular  hexagon  is  produced,  show  that 
the  exterior  angle  is  equal  to  the  angle  of  an  equilateral 
triangle. 

3.  The  exterior  angle  of  a  regular  polygon  is  one-fifth  of 
a  right  angle  :  find  the  number  of  sides  in  the  polygon. 

4.  The  interior  angle  of  a  regular  polygon  is  five-thirds 
of  a  right  angle:  find  the  number  of  sides.in  the  polygon. 

5.  The  side  AB  of  the  triangle  ABO  is  produced  to  D, 
so  that  BD  is  equal  to  BO  :  prove  that  the  angle  ABO  is 
double  the  angle  ADO. 

6.  How  many  sides  has  a  polygon,  the  sum  of  whose  in- 
terior angles  is  four  times  that  of  its  exterior  angles  ? 

/  O 


56 


PLANE  GEOMETBY. 


I 


M 


Miscellaneous  Theorems. 
Proposition  38.    Theorem. 

152.  Jf  three  or  more  parallels  intercept  equal  lengths 
on  any  transversal,  they  intercept  equal  lengths  on  every 
transversal. 

Hyp.  Let  the  ||  s  AE,  BF,  CG,  DH, 
cut  the  two  transversals  MN  and  OP 
at  the  pts.  A,  B,  0,  D,  and  E,  F,  G,  H, 
so  that 

AB  =  BC  =  CD  =  etc. 
To  prove      EF  =  FG  =  GH  ==  etc. 
Proof.    From  E,  F,  G,  draw  EK, 
FL,  GS,  all  II  to  MN. 

Then 


A 

\e 

B 

P\f 

■= 

K    \ 

J 

I\h 

EK  =  AB,  FL  =  BO,  GS  =  CD, 
hdng  opp.  sides  of  a  r'7(129). 


(Ax.  1) 


Also 
and 


.•.EK  =  FL=GS  =  etc. 
ZFEK=ZGFL  =  ZHGS, 

Z  EFK  =  Z  FGL  =  Z  GHS, 

being  ext.-int.  AsofW  lines  (77). 

.-.  aEFK  =  aFGL=aGHS, 

Mmng  a  side  and  the  two  adj.  As  equal,  each  to  each  (100,  105). 

.•.EF  =  FG  =  GH, 

being  homologous  sides  of  equal  as  (109).  q.e.d. 

153.  Cor.  Since  FK  =  GL  =  HS, 
.•.BF-AE=:FK, 
and  CG  -  BF  =  GL  =  FK. 

Therefore,  the  intercepfted  p^drt  of  each  jMrallel  ivill  differ 
in  length  from  the  next  intercept  hy  the  same  amount. 


I 


BOOK  L-MISCELLANEOVS  THEOllEMS.  57 

Proposition  39.    Theorem. 

154.  The  straight  line  drawn  through  the  middle  point 
of  a  side  of  a  triangle  parallel  to  the  base,  bisects  the  re- 
maining  side,  and  is  equal  to  half  the  base. 

Hyp.  Let  ABC  be  a  A,  D  the  middle 

poiut  of  AB,  and  DE  a  line  ||  to  BC,  meet-  A 

ing  AC  at  E.  /\ 

To  prove  (1)                 AE  =  EC  ;  o/- \e 

and          (2)                  DE  =  iBC.  /               \ 

Proof      (1)  Through  A  draw  a  line  ||  b                       C 

to  BC.  .    ~ 

Then,  because  AD  =  DB,     ^4^  '^ jli^(Hyp.) 

...AE^EC.    ^^^^^      €C 
if  lis  intercept  equal  lengths  on  one  transversal,  they  intercept  equal 
lengths  on  every  transversal  (152). 

(2)  The  intercepted  part  BC  exceeds  DE  as  much  as  DE 
exceeds  the  intercepted  part  at  A  (153).  But  this  latter 
intercept  =  0. 

.•.BC-DE  =  DE. 

.-.   DE^^BC.  Q.E.D. 

155.  Cor.  1.  The  line  joini7ig  the  middle  points  of  two 
sides  of  a  triangle  is  parallel  to  the  third  side,  and  equal  to 
half  of  it, 

156.  Cor.  2.  Let  ABCD  be  a  trape-        Dy ^C 

zoid,'  AG  =:  CD,  and  GH  |1  to  AB.  ^ 

Then    BH  =  HC,    for,    since    AB, 

GH,  and  DC  are  parallels  intercept-  A''^ ^B 

ing  equal  lengths  AG  and  GD  on  the 

transversal   AD,   they  intercept    equal  lengths  on   every 

transversal  (1^^) 

Also  -       AB  -  GH  =  GH -DC.  (153) 

.•.GH==:i(AB  +  DC). 

Therefore,  the  line  drawn  through  the  middle  point  of  one 
of  the  non-parallel  sides  of  a  trapezoid  parallel  to  the  bases, 
bisects  the  opposite  side,  and  is  equal  to  half  their  sum. 


58  PLANE  GEOMETRY, 

The  Locus  of  a  Poij^^t. 

157.  The  locus  of  a  point  is  the  line,  or  system  of  lines, 
which  contains  all  the  points,  and  only  those  points,  that 
satisfy  a  given  condition. 

A  locus  is  sometimes  defined  as  the  path  traced  out  by 
a  point  which  moves  in  accordance  with  a  given  law. 

Thus,  the  locus  of  a  point  which  is  always  at  a  given  dis- 
tance from  a  given  straight  line,  is  a  pair  of  straight  lines. 

158.  In  order  to  infer  that  a  certain  line,  or  system  of 
lines,  is  the  locus  of  a  point  under  a  given  condition,  it  is 
necessary  to  prove,  (1)  that  any  point  which  fulfils  the 
given  condition  is  on  the  supposed  locus  ;  and  (2)  that  every 
point  on  the  supposed  locus  satisfies  the  give?i  condition, 

159.  T7ie  locus  of  a  point  tvhich  is  always  equidistant 
from  two  given  points,  is  the  perpendic^dar  bisector  of  the 
line  joining  them  (67). 

EXERCISE. 

Find  the  locus  of  the  middle  point  of  a  straight  line 
drawn  from  a  given  point  to  meet  a  given  straight  line  of 
unlimited  length. 

Let  A  be  the  given  pt.,  and  BC  the  ^  !p  — /b  x'E  C 
given  st.  line.  _±LL__Pi'jQrl!^ 

(1)  Let  AD  be   any  straight  line         i^^^"'' 
drawn  from  A  to  meet  BC,  and  let  P       p^ 

be  its  mid.  pt. 

Draw  AF  J.  to  BC,  and  bisect  AF  at  H. 

Join   HP,  and  produce  it  indefinitely. 

Then  HP  is  ||  to  FD.  (155) 

.  • .  P  is  on  the  st.  line  which  passes  through  the  fixed  pt. 
Hand  is  II  to  BC. 

(2)  Every  pt.  in  HP,  or  HP  produced,  satisfies  the  re- 
quired condition. 

For  in  this  st.   line  take  any  pt.  Q.     Join  AQ,  and 

produce  it  to  meet  BC  in  E. 
Then  Q  is  the  mid.  pt.  of  AE.  (154) 

Hence  the  required  locus  is  the  st.  line  ||  to  BC  and  pass- 
ing through  the  mid.  pt.  of  the  J_  from  A  to  BO. 


BOOK  L— MISCELLANEOUS  THEOREMS.  59 


Proposition  40.    Theorem. 

160.  (1)  Every  point  in  the  bisector  of  an  angle  is  equally 
11  distant  from  the  sides  of  the  angle;   and  (2)  conversely, 

every  point  luithin  an  angle,  and  equally  distant  from  its 
sides,  is  in  the  bisector  of  the  angle.  ^C 

(1)  Hyp,  Let  AE  be  the  bisector  of 
the  Z  BAG  ;  P  any  point  in  AE  ;  and  /^^^    ^^^E 
PD,  PH  J_s  to  AB,  AC. 

To  prove  PD  =  PH.  (59)    f 

Proof  Because  in  the  rt.  A  s  APD^  APH, 

ZDAP  =  ZHAP,  (Hyp.) 

and  AP  is  common, 

.-.  aAPD=a  APH, 
having  the  hypotenuse  and  an  acute  Z  equal  in  each  (106). 

.•.PD  =  PH. 

being  homologous  sides  of  equal  as  (109). 

(2)  Hyp,  Let  PDi.to  AB  =  PH  _L  to  AC. 

To  prove  that  P  is  on  the  bisector  of  Z  BAO. 

Proof  Join  PA. 
Then  A  APD  =  A  APH, 

having  the  hypotenuse  and  a  side  equal  in  each  (110). 
.-.  Z  PAD=  Z  PAH, 

teing  homologous  As  of  equal  As  (109). 

.  • .  P  is  on  the  bisector  of  the  angle  BAG.     q.e.d. 

161.  GoR.  ^very  point  within  the  angle,  but  not  on  the 

bisector,  is  unequally  distant  from  the  tivo  sides, 

162.  ScH.  The  bisector  of  an  angle  is  the  locus  of  all  the 
points  situated  within  the  angle^  which  are  eq^ially  distant 
from  its  sides,  y' 


60  PLANE  GEOMETRY. 


The  Concuerekce  of  Straight  Lines  iif  a 

TRIAiq^GLE. 

163.  Three  or  more  straight  lines  are  said  to  be  concur- 
rent when  they  meet  in  a  point. 

164.  Three  or  more  points  are  said  to  be  coUitiear  when 
they  lie  upon  one  straight  line. 

165.  A  straight  line  from  any  vertex  of  a  triangle  to  the 
middle. point  of  the  opposite  side  is  called  a,  medial,  or  a 
7nedian,  of  the  triangle. 

Proposition  41 .    Theorem. 

166.  The  bisectors  of  the  three  angles  of  a  triangle  are 
concurrent. 

Hyp.  Let  ABC  be  a  A ;  0 A,  OB, 
DC,  the  bisectors  of  the  Z  s  A,  B, 
and  0. 

To  prove  that  these  bisectors 
meet  in  a  pt. 

Proof  Let  the  bisectors   OA, 
OB  meet  at  0. 
Draw  the  _Ls  OL,  OM,  ON. 

Because  0  is  on  the  bisector  of  /  BAC, 

.  • .  OL  =  OK  (160) 

Because  0  is  on  the  bisector  of  Z  ABC, 

.•.OL=:OM.  (160) 

.•.OK  =  OM.  (Ax.  1) 

.*.  0  is  on  the  bisector  of  the  Z  ACB, 
heing  equally  distant  from  its  sides  [160  (2)], 
that  is,  0  must  lie  on  the  bisector  CD. 
.  • .  the  bisectors  of  the  three  Z  s  meet  at  the  point  0. 

Q.E.D. 

167.  Cor.  TJie  point  0  is  equalUj  distant  from  the  three 
sides  of  the  triangle. 


BOOK  L— MISCELLANEOUS  THEOBEMS. 


61 


Proposition  42.    Theorem. 

168.  TJie  perpendicular  Msectors  of  the  three  sides  of  a 
triangle  are  concurreiit. 

Hyp,  Let  ABC  be  a  A  ;  D,  E,  F, 
the  middle  pts.  of  its  sides ;  and 
DO,  EO,  EH,  the  i_s  erected  at 
D,  E,  F. 

To  prove  that  these  J_s  meet  in 
a  pt. 

Proof  The  two  ±s  OD,  OE, 
since  they  cannot  be  1| ,  will  meet  at  the  pt.  0. 

Join  OA,  OB,  OC. 

Because  0  is  in  the  _L  bisector  OD, 

.•.OA  =  OB.  (66) 

Because  0  is  in  the  _L  bisector  OE, 

. -.06  =  00.  {m) 

.-.OArrOC.  (Ax.  1) 

.-.  0  is  in  the  1  bisector  of  AC;      (67) 

that  is,  0  must  lie  on  the  J_  bisector  HE. 

.  • .  the  three  J.  bisectors  meet  at  the  pt.  0.      q.e.d. 

169.  Cor.  The  point  of  intersection  of  the  perpendicular 
bisectors,  is  equally  distant  from  the  three  vertices  of  the 
triangle. 


EXERCISES. 

1.  If  the  1;riangle  in  Prop.  41  is  equilateral,  find  the  value 
of  the  angle  AOB. 

2.  If  the  same  triangle  is  isosceles,  and  the  angle  0  is 
three  times  as  great  as  either  of  the  angles  A  and  B,  find 
the  value  of  the  angle  AOB. 


62 


PLANE  GEOMETRY, 


Proposition  43.    Theorem. 

170.  The  peiyendiculars  from  the  vertices  of  a  triangle 
to  the  opposite  sides  are  concurrent. 
Hyp.  Let  ABC  be  a  A,  and  p^- -^ -^^q 


AD,  BE,  CF  the  three  J_s  from 
A,  B,  C  to  the  opp.  sides. 

To  provg  that  these  J_s  meet 
ill  a  pt. 

Proof.  Through  -A,  B,  0, 
draw  RQ,  RP,  PQ  ||  respectively 
to  BC,  AC,  AB. 

Then,  since  ABPC  is  a  OJ, 

^ 

A 

\ 

v: 

(124) 

.-.  BP  = 

AC. 

(139) 

And  since  ARBC  is  a  CD, 

(124) 

.-.  RB  = 

AC. 

(129) 

.-.  RB  = 

BP. 

(Ax.l) 

Similarly,  RA  =  AQ,  and  PC  =  CQ; 

that  is,  A,  B,  C,  are  the  mid.  pts.  of  QR,  RP,  PQ. 

Since  AC  is  |1  to  PR,  and  BE  is  J_  to  AC, 

.-.  BE  is  also  i.  to  PR. 
A  line  1  to  one  of  two  Wsis  i.  to  the  other  (71). 

Similarly,  AD  and  CF  are  JL  to  RQ  and  PQ. 

.  * .  these  three  J_s  meet  in  a  pt., 
leing  tlie  JL  bisectors  oftlie  three  sides  oftJie  A  PQR  (168).  q.e.d. 

171.  Def.  The  intersection  of  the  perpendiculars  from 
the  vertices  of  a  triangle  to  the  opposite  sides  is  called  its 
orthocentre. 

The  triangle  formed  by  joining  the  feet  of  the  perpen- 
diculars is  called  the  pedal  or  orthoQentric  triangle. 


BOOK  1. -MISCELLANEOUS  THEOREMS 


63 


-^ 


Proposition  44.    Theorem. 

172.  The  three  medial  lines  of  a  triangle  are  concurrent 
in  a  point  of  trisection,*  the  greater  segment  in  each  being 
toiuards  the  angular  point. 

Hijp.  Let  ABO  be  a  A  ;  D,  E,  F, 
the  mid.  pts.  of  BO,  AO,  AB;  and 
AD,  BE,  OF,  the  three  medial 
lines. 

To  prove  that  these  lines  meet  in 
the  pt.  that  trisects  them. 

Proof,  Let  the  two  medials  BE 
and  OF  meet  in  0.     Bisect  BO  in  H,  and  00  in  K. 

Join  HK,  KE,  EF,  FH. 

In  the  A  BOO,  because  H  and  K  are  the  mid.  pts.  of  BO 
and  00, 

.-.  HK  is  II  and  =  to  J  BO.  (155) 

Also,  in  thfe  A  ABO,  because  E  and  F  are  the  mid.  pts. 
of  AO  and  AB, 

.-.  EF  is  II  and  =  to  I  BO.  (155) 

.-.  EFHK  is  a  OJ, 

having  two  opp.  sides  equal  and  \\  (133). 

.-.  0  is  the  mid.  pt.  of  Ell  and  FK. 
The  diagonals  ofacj  bisect  each  other  (134). 
.-.  EO  =  OH  =  HB,  and  FO  =  OK  =  KO; 
or  EO  --=  J-  EB,  and  FO  =  i  FO. 
That  is,  the  medial  BE  cuts  the  medial  OF  at  a  pt.  0, 
one- third  the  way  from  F  to  0. 

In  the  same  way  it  may  be  proved  that  the  medial  AD 
cuts  the  medial  OF  at  a  pt.  one-tliird  the  way  from  F  to  0; 
that  is,  at  the  same  pt.  0. 

.'.  The  three^medials  are  concurrent  in  the  point  that  tri- 
sects them.  q.p:.d. 

Note.— The  point  of  intersection  of  the  three  medians  of  a  triangle  is  called 
the  centroid. 


*  When  a  line  is  divided  into  three  equal  parts  it  is  said  to  be  trisected, 


64 


PLANE  GEOMETRY. 


Symmetry. 


Symmetry  with  ref^pect  to  an  axis. 


\P 


IvT 


173.  Two  points  are  said  to  be  symmetrical  with  respect 
to  a  straight  line,  when  the  straight  line  bisects  at  right 
angles  tli^  straight  line  joining  the  two 
points.  . 

Thus,  the  two  points  P  and  P'  are 
symmetrical  with  respect  to  the  line 
MN,  if  MN  bisects  PP'  at  right  angles. 

The  straight  line  MN  is  called  the 
axis  of  symmetry. 

If  we  take  the  plane  containing  the 
pt.  P,  and  turn  it  about  the  axis  MN,  until  the  upper  part  is 
brought  down  on  the  part  below  MN,  the  line  AP  will  take 
the  direction  AP',  and  the  point  P  will  coincide  with  the 
point  P'.  Thus,  when  two  points  are  symmetrical  with  re- 
spect to  an  axis,  if  one  of  the  parts  of  the  plane  be  revolved 
about  the  axis  to  bring  ]t  down  on  the  other  part,  the 
symmetrical  points  coincide. 

C 


ip' 


1 74.  Two  figures  are  said  to  be  symmetrical  with  re- 
spect to  an  axis,  when  every  point  in  one  figure  has  its 
symmetrical  point  in  the  other. 

Thus,  the  figures  ABC,  A'B'C  are  symmetrical  with 
respect  to  the  axis  MN,  if  every  point  in  the  figure  ABC 
has  a  symmetrical  point  in  A'B'C  with  respect  to  MN. 

In  all  cases,  two  figures  that  are  symmetrical  with  re- 


BOOK  I.—SYMMETliY. 


65 


II 

I 


spect  to  an  axis,  can  be  applied  one  to  the  other,  by  revolv- 
ing either  about  the  axis;  consequently  they  are  equal. 
The  corresponding  symmetrical  lines  of  symmetrical  fig- 
res  are  called  Jionioloc/ous  lines.  Thus,  in  the  symmetric- 
1  figures,  ABC,  A'B'C,  the  homologous  lines  are  AB 
ind  A'B',  BC  and  B'C,  AC  and  A'C 


Symmetry  luith  respect  to  a  point. 


175.  Two  points  are  said  to  \>q  symmetrical  vf\i\\YQ^^eci 
to  a  third  point,  when  this  third  point  bisects  the  straight 
line  joining  the  two  points. 

Thus,  P  and  P'  are  symmetrical  with 
respect  to  A,  if  the  straight  line  PP'  is  P  A  P 

bisected  at  A. 

The  point  A  is  called  the  centre  of  symmetry, 

116.  Two  figures  are  said  to  be  symmetrical  with  re- 
spect to  a  centre,  when  every  point  in  one  figure  has  its 
symmetrical  point  in  the  other. 

Thus,  the  figures  ABC,  A'B'C 
are  symmetrical  with  respect  to 
the  centre  0,  if  every  point  in  the 
figure  ABC  has  a  symmetrical 
point   in  A'B'C. 

177.  A  figure  is  symmetrical 
with  respect  to  an  axis,  when  it 
can  be  divided  by  that  axis  into  two 
figures  symmetrical  with  respect 
to  the  axis. 

A  figure  is  symmetrical  with  re- 
spect to  a  centre,  when  every 
straight  line  drawn  through  that 
centre  cuts  the  figure  in  two  points 
symmetrical  with  respect  to  this 
centre. 


66  PLANE  GEOMETRY. 

Proposition  45.  Theorem. 

178.  If  a  figure  is  symmetrical  with  respect  to  two  axes 
at  right  angles  to  each  other,  it  is  also  symmetrical  with 
respect  to  their  intersection  as  a  centre. 


P  G 

Y 

F     _. 

f^ 

-  -^->9 

H 

^^^ —  -y 

E 

^" 

0 

X 

P    B 

Y'° 

Hyp,  Let  the  figure  ABCDEFGH  be  symmetrical  with 
respect  to  the  two  _L  axes  XX',  YY',  which  intersect  at  0. 
To  prove  that  0  is  the  centre  of  symmetry  of  the  figure. 
Proof.  Let  P  be  any  pt.  in  the  perimeter  of  the  figure. 
Draw  PRP'  J_  to  XX',  and  PSQ  J.  to  YY'. 
Join  RS,  OP',  and  OQ. 

Then,  because  the  figure  is  symmetrical  with  respect  to 
XX', 

.-.  PR  :=  P'R. 
And  since  PR  =  OS, 

.-.  P'R  =  OS. 
.-.  RP'0SisaZl7, 
Jiatring  two  opp.  sides  =  and  ||  (133). 

.-.  RSis  =  and  ||  to  P'O, 
being  opp.  sides  of  a  CJ  (124). 
Also,  since  the  figure  is  symmetrical  with  respect  to  YY', 

.-.PS^SQ. 
And  since  PS  =  OR,  .-.  SQ  =  OR. 

.  •.  SROQ  is  a  /=7;  .  • .  RS  is  =  and  1|  to  OQ. 
Because  both  P'O  and  OQ  are  =  and  ||  to  RS, 
.  • .  the  points  P',  0,  Q  are  in  the  same  st.  line  which  is 
bisected  at  0. 

.*.  any  st.  line  P'OQ,  drawn  through  0,  is  bisected  at  0. 
.  • .  the  figure  is  symmetrical  with  respect  to  0  as  a  cen- 
tre (177).  9.E.P, 


BOOK  L— EXERCISES.  67 

EXERCISES. 

The  following  theorems  are  given  for  the  exercise  of  the 
student,  that  he  may  work  out  his  own  demonstrations. 
The  student  can  make  no  solid  acquisitions  in  geometry, 
without  frequent  practice  in  the  application  of  the  princi- 
ples he  has  acquired.  lie  should  not  merely  learn  the 
demonstration  of  a  number  of  theorems,  but  he  should  ac- 
quire the  power  of  grasping  and  demonstrating  geometric 
theorems  for  himself,  and  this  power  can  never  be  gained 
by  memorizing  demonstrations.  He  should  understand 
that  the  ability  to  investigate,  to  reason  for  himself,  is  the 
chief  object  for  the  attainment  of  which  he  should  strive. 
Diligent  application,  systematic  practice  in  devising  proofs 
of  new  propositions,  is  indispensable. 

In  the  process  of  finding  a  demonstration  the  student 
should  first  construct  a  diagram,  and  state  the  hypothesis, 
including  in  the  statement  not  only  what  the  theorem  says, 
but  what  it  implies.  He  should  also  examine  the  conclu- 
sion, and  see  what  it  says  and  what  it  implies,  and  discover 
the  relation  between  the  hypothesis  and  the  conclusion.  A 
correct  diagram  is  most  useful  in  suggesting  the  steps  by 
which  a  theorem  is  to  be  demonstrated.  If  the  student 
will  ask  himself  why  he  takes  any  particular  step,  he  may 
avoid  the  habit  of  random  guessing,  and  with  more  cer- 
tainty discover  the  correct  and  direct  process  for  effecting 
the  demonstration.  Sometimes  it  will  be  necessary  to  draw 
additional  lines  in  the  diagram,  and  to  call  to  mind  the 
different  theorems  which  apply  to  the  figure  thus  formed. 

The  demonsti-ation  must  be  framed  in  the  simplest 
manner,  but  without  omitting  any  logical  step.  This  is  a 
matter  of  practice,  in  which  no  general  rule  can  be  given. 

The  student  should  express  each  step  of  the  demonstra- 
tion completely  and  fully.  Tlie  most  common  fault  is  that 
of  passing  over  steps  in  the  demonstration  because  the  con- 
clusion seems  to  be  obvious, 


68  PLANE  QEOMETllT. 

It  is  often  the  case  that  the  clearness  of  intuition  acquired 
by  a  practised  geometrician  will  make  him  impatient  of  the 
successive  steps  of  detailed  reasoning,  and  he  will  be  eager 
to  conduct  his  pupil  to  the  desired  end  by  a  shorter  and  an 
easier  road.  But  it  is  a  great  misfortune  to  the  learner  if 
he  is  deprived  of  that  useful  disciplme  in  geometric  reason- 
ing which,  however  tedious  it  may  seem  in  its  application 
to  short  and  easy  propositions,  is  indispensable  to  the  in- 
vestigation of  more  lengthy  and  difficult  ones. 

One  of  the  great  objects  of  the  study  of  geometry  is  to  cultivate  the  habit  of 
examining  the  logical  foundations  of  those  conclusions  which  are  accepted  with- 
out critical  examination.  The  feeling  of  security  that  a  conclusion  is  right  before 
its  foundation  has  been  examined  is  a  most  fruitful  source  of  erroneous  opinions, 
and  the  person  who  neglects  the  habit  of  inquiring  into  what  appears  obvious 
is  liable  to  pass  over  things  which,  had  they  been  carefully  examined,  would  have 
changed  the  conclusion.* 

1.  If  the  angles  ABC  and  ACB  at  the  base  of  an  isosceles 
triangle  be  bisected  by  the  lines  BD,  CD,  show  that  DBC 
will  be  an  isosceles  triangle. 

2.  BAC  is  a  triangle  having  the  angle  B  double  the  angle 
A.  If  BD  bisects  the  angle  B  and  meets  AC  at  D,  show 
that  BD  is  equal  to  AD. 

3.  In  the  triangle  ABC,  the  angle  A  =  50°,  the  angle 
B  =  70°.  What  angle  will  the  bisectors  of  these  two  angles 
make  with  each  other? 

4.  In  the  preceding  triangle,  what  will  be  the  values  of 
the  three  exterior  angles  ? 

5.  A  given  angle  BAG  is  bisected ;  if  CA  is  produced  to 
G,  and  the  angle  BAG  bisected,  prove  that  the  two  bisect- 
ing lines  are  at  right  angles  to  each  other. 

6.  ACB,  ADB  are  two  triangles  on  the  same  side  of  AB, 
such  that  AC  =  BD,  and  AD  =  BC,  and  AD  and  BC  in- 
tersect at  0 :  prove  that  the  triangle  AOB  is  isosceles. 

7.  ABC  is  a  triangle,  and  the  angle  A  is  bisected  by  a 
line  which  meets  BC  at  D:  show  that  BA  is  greater  than 
BD,  and  CA  is  greater  than  CD. 

*  Newcomb. 


BOOK  L— EXERCISES.  69 

8.  If  one  angle  of  a  triangle  is  equal  to  the  sum  of  the 
(  other  two,  the  triangle  can  be  divided  into  two  isosceles 

triangles. 

9.  If  the  angle  C  of  a  triangle  is  equal  to  the  sum  of  the 
angles  A  and  B,  the  side  AB  is  equal  to  twice  the  line  join- 
ing C  to  the  middle  point  of  AB. 

10.  A  line  bisects  the  angle  A  of  a  triangle  ABC  ;  from 
B  a  perpendicular  is  drawn  to  this  bisecting  line,  meeting 
it  at  D,  and  BD  is  produced  to  meet  AC  or  AC  produced 
at  E :  show  that  BD  =  DE. 

11.  A  line  is  drawn  terminated  by  two  parallel  lines; 
through  its  middle  point  any  line  is  drawn  and  terminated 
by  the  parallel  lines.  Show  that  the  second  line  is  bisected 
at  the  middle  point  of  the  first. 

12.  If  through  any  point  equidistant  from  two  parallel 
lines,  two  lines  be  drawn  cutting  the  parallel  lines,  they 
will  intercept  equal  portions  of  these  parallel  lines. 

13.  If  the  line  bisecting  the  exterior  angle  of  a  triangle 
be  parallel  to  the  base,  show  that  the  triangle  is  isosceles. 

14.  If  a  line  be  drawn  bisecting  one  of  the  angles  of  a 
triangle  to  meet  the  opposite  side,  the  lines  drawn  from  the 
point  of  intersection  parallel  to  the  other  sides,  and  termi- 
nated by  these  sides,  will  be  equal. 

Let  ABC  be  the  a  ;  let  a  line  be  drawn  bisecting  z  A  and  meeting  BC  at  D,  etc. 

15.  The  side  BC  of  a  triangle  ABC  is  produced  to  a  point 
D;  the  angle  ACB  is  bisected  by  the  line  CE  which  meets 
AB  at  E.  A  line  is  drawn  through  E  parallel  to  BC,  meet- 
ing AC  at  F,  and  the  line  bisecting  the  exterior  angle  ACD 
at  G.     Show  that  EF  =  FG. 

16.  A  line  drawn  at  right  angles  to  BC,  the  base  of  an 
isosceles  triangle  ABC,  cuts  the  side  AB  at  D  and  CA  pro- 
duced at  E:  show  that  AED  is  an  isosceles  triangle. 

From  A  draw  a  line  bisecting  ZBAC,  and  meeting  BC  at  F,  etc. 


70  PLANS!  omMETRT. 

17.  If  the  lines  bisecting  the  angles  at  the  base  of  an 
isosceles  triangle  be  produced  to  meet,  they  will  contain  an 
angle  equal  to  an  exterior  angle  of  the  triangle. 
''18.  A  is  the  vertex  of  an  isosceles  triangle  ABC,  and  BA 
i^  produced  to  D,  so  that  AD  =  BA;  and  DC  is  drawn: 
show  that  BCD  is  a  right  angle. 

•^OtMl9.  ABC  is  a  triangle,  and  the  exterior  angles  at  B  and 
..^.  are  bisected  by  the  lines  BD,  CD  respectively,  meeting  at 
, p.;  show  that  the  angle  BDC  together  with  half  the  angle 
BAC  make  up  a  right  angle. 

20.  In  the  triangle  ABC  the  side  BC  is  bisected  at  E, 
and  AB  at  G;  AE  is  produced  to  F  so  that  EF  =:  AE,  and 
CG  is  produced  to  H  so  that  GH  ==  CG:  show  that  FB  and 
HB  are  in  one  straight  line. 

21.'  Lines  are  drawn  through  the  extremities  of  the  base 
of  an  isosceles  triangle,  making  angles  with  it  on  the  side 
remote  from  the  vertex,  each  equal  to  one-third  of  one  of 
the  equal  angles  of  the  triangle  and  meeting  the  sides  'pro- 
duced :  show  that  the  three  triangles  thus  formed  are 
isosceles. 

22.  AEB,  CED  are  two  lines  intersecting  at  E  ;  lines  AC, 
DB  are  drawn  forming  two  triangles  ACE,  BED;  the 
angles  ACE,  DBE  are  bisected  by  the  lines  CF,  BF,  meet- 
ing at  F.  Show  that  the  angle  CFB  is  equal  to  half  the 
sum  of  the  angles  EAC,  EDB. 

I  AEC  =  I  ECB  +  I  EBC,  etc. 

23.  If  a  quadrilateral  have  two  of  its  opposite  sides 
parallel,  and  the  other  two  equal  but  not  parallel,  any 
two  of  its  opposite  angles  are  together  equal  to  two  right 
angles. 

24.  On  the  sides  AB,  BC,  and  CD  of  a  parallelogram 
ABCD  three  equilateral  triangles  are  described,  that  on  BC 
towards  the  same  parts  as  the  parallelogram,  and  those  on 
AB,  CD  towards  the  opposite  parts:  show  that  the  dis- 
tances of  the  vertices  of  the  trianorles  on  AB,  CD  from  that 


BOOK  I.-EXERCI8E8.  71 

on  BC  are  respectively  equal  to  the  two  diagonals  of  the 
panillelogram. 

25.  A,  B,  C  are  three  points  in  a  straight  line,  such  that 
AB  =  BC:  show  that  the  sum  of  the  perpendiculars  from 
A  and  C  on  any  straight  line  which  does  not  pass  between 
A  and  0  is  double  the  perpendicular  from  B  on  the  same 
line. 

26.  In  a  right  triangle,  show  that  the  medial  drawn 
from  the  right  angle  is  equal  to  half  the  hypotenuse. 

27.  If  P  and  Q  are  the  feet  of  the  perpendiculars  from 
A  on  the  lines  bisecting  the  angles  B  and  C  of  a  triangle 
ABC ;  show  that  PQ  is  parallel  to  BC. 

Produce  AP,  AQ  to  meet  BC  in  D,  E,  etc. 

28.  AD,  BE,  CF,  the  perpendiculars  from  the  vertices 
of  a  triangle  ABC,  intersect  in  0  ;  prove  that  the  angle 
BOF  =  the  angle  BAC,  the  angle  FO A  =  the  angle  ABC, 
and  the  angle  BOD  =  the  angle  BCA. 

29.  What  is  the  magnitude  of  each  angle  of  the  follow- 
ing regular  figures  ?  a  pentagon,  an  octagon,  a  decagon. 

30.  From  the  angle  A  of  a  triangle  ABC  a  perpendicu- 
lar is  drawn  to  the  opposite  side,  meeting  it,  produced  if 
necessary,  at  D;  from  the  angle  B  a  j)erpendicular  is  drawn 
to  the.  opposite  side,  meeting  it,  produced  if  necessary,  at 
E:  show-that  the  lines  which  join  D  and  E  to  the  middle 
point  of  AB  are  equal. 

31.  From  the  angles  at  the  base  of  a  triangle  perpen- 
diculars are  drawn  to  the  opposite  sides,  produced  if  neces- 
sary: show  that  the  line  joining  the  points  of  intersection 
will  be  bisected  by  a  perpendicular  drawn  to  it  from  the 
middle  point  of  the  base. 

32.  The  lines  which  join  the  middle  points  of  adjacent 
sides  of  any  quadrilateral,  form  a  parallelogram. 

33.  The  sides  AB,  AC  of  the  triang'e  ABC  are  bisected 
in  D  and  E  respectively;  BE  and  CD  are  produced  to  F 


72  PLANE  GEOMETRY. 

and  G,  so  that  EF  =  BE  and  DG  =  DO :  prove  that  F,  A, 
G  are  collinear. 

34.  ABODE  is  a  regular  pentagon;  AO,  AD  are  joined: 
prove  that  each  of  the  angles  AOD,  ADO  is  double  the 
angle  OAD. 

Draw  DF  II  to  AE,  F  being  in  AC,  etc. 

35.  Two  medians  of  a  triangle  are  equal :  prove  (with- 
out assuming  that  they  trisect  each  other)  that  the  triangle 
is  isosceles. 

36.  ABCD,  BAOE  are  parallelograms  on  the  same  baso 
AB,  such  that  the  diagonals  BD,  AM  are  equal :  prove  that 
the  other  diagonals  AO,  BO  are  equal. 

37.  ABOD  is  a  parallelogram;  E  is  the  middle  point  of 
BO;  AB  and  DE  produced  meet  in  F:  prove  that  the 
triangle  DBF  is  half  the  parallelogram  ABOD. 

38.  If  two  right  triangles  ABO,  ABD  be  on  the  same 
hypotenuse  AB,  and  the  vertices  0  and  D  be  joined,  the 
pair  of  angles  subtended  by  any  side  of  the  quadrilateral 
thus  formed  are  equal. 

39.  The  angles  made  with  the  base  of  an  isosceles  tri- 
angle by  perpendiculars  from  its  extremities  on  the  equal 
sides  are  each  equal  to  half  the  vertical  angle. 

40.  The  angle  included  between  the  internal  bisector  of 
one  base  angle  of  a  triangle  and  the  external  bisector  of  the 
other  base  angle  is  equal  to  half  the  vertical  angle. 

41.  The  sum  of  the  distances  of  any  point  m  the  base  of 
an  isosceles  triangle  from  the  equal  sides  is  equal  to  the  dis- 
tance of  either  extremity  of  the  base  from  the  opposite  side. 

42.  The  sum  of  the  perpendiculars  from  any  point  in 
the  interior  of  an  equilateral  triangle  is  equal  to  the  per- 
pendicular from  any  vertex  on  the  opposite  side. 

43.  AD  and  BO  are  two  parallel  lines  cut  obliquely  by 
AB,  and  perpendicularly  by  AO  ;  and  between  these  lines 
we  draw  BED,  cutting  AO  in  E,  such  that  ED  =  2AB  ; 
prove  that  the  angle  DBO  is  one-third  of  ABO. 

44.  If  0  be  the  point  of  concurrence  of  the  bisectors  of 


BOOK  I— EXERCISES.  73 

the  angles  of  the  triangle  ABC,  and  if  AO  produced  meet 
BC  in  D,  and  from  0,  OE  be  drawn  perpendicular  to  BC  ; 
prove  that  the  angle  BOD  =  the  angle  COE. 

45.  The  bisectors  of  the  external  angles  of  a  quadrilat- 
eral form  a  circumscribed  quadrilateral  the  sum  of  whose 
opposite  angles  is  equal  to  two  right  angles. 

4G.  The  locus  of  a  point  equidistant  from  two  given  in- 
tersecting lines  is  two  lines  at  right  angles  to  each  other. 

Bisect  the  adj.  Zs  between  the  given  Unes,  etc. 

47.  Any  line  is  drawn  cutting  two  fixed  intersecting  lines, 
nd  the  two  angles  on  the  same  side  of  it  are  bisected :   find 

-the  locus  of  the  point  of  intersection  of  the  bisecting  lines. 

48.  The  leaf  of  a  book  is  turned  down  so  that  the  corner 
always  lies  on  the  same  line  of  printing  :  find  the  locus  of 
the  foot  of  the  perpendicular  from  the  corner  to  the  crease. 

49.  ABO  is  a  triangle,  D  is  the  middle  point  of  BO,  E 
the  middle  point  of  AD  ;  let  BE  produced  meet  AO  in  F  : 
prove  that  AO  is  trisected  in  F. 

Draw  DG  II  to  BF  meeting  AC  in  G,  etc. 

50.  D,  E,  F  are  the  middle  points  of  the  sides  of  a  tri- 
angle, N  is  the  foot  of  the  perpendicular  from  the  angle 
opposite  D  to  the  side  which  D  bisects  :  prove  that  z  EDF 
=  Z  ENF. 

Let  D,  E,  F  lie  in  BC,  CA,  AB  respectively;  ED  is  II  to  AB,and  DF  II  to  AC,  etc. 

51.  If  A,  B,  0  denote  the  angles  of.  a  A,  prove  that 
i(A  +  B),  i(B  +  0),  i(0  + A)  will  be  the  angles  of  a  A 
formed  by  any  side  and  the  bisectors  of  the  external  angles 
between  that  side  and  the  other  sides  produced. 

52.  If  the  exterior  angles  of  a  A  be  bisected,  the  three 
external  As  formed  on  the  sides  of  the  original  A  are  equi- 
angular. 

53.  If  a  hexagon  have  its  opposite  sides  equal  and  paral- 
lel, the  three  straight  lines  joining  the  opposite  angles  are 
concurrent. 

54.  Show  that  a  parallelogram  is  symmetrical  with  respect 
to  its  centre. 


Book   II. 
THE  CIKCLE. 


Definitions. 

179.  A  circle  is  a  plane  figure  bounded  by  a  curved  line 
called  the  circiimfereilce,  every  point  of  which  is  equally 
distant  from  a  point  within  called  the  centre. 

A  radius  is  a  straight  line  drawn  from  the  centre  to  the 
circumference. 

A  diameter  is  a  straight  line  drawn  through  the  centre, 
and  terminated  both  ways  by  the  circumference. 

Thus,  in  the  figure,  ABODE  is  the 
circumference;  the  space  included 
within  the  circumference  is  the 
circle;  0  is  the  centre;  OA,  OB,  00, 
are  radii;  AOO  is  a  diameter. 

From  the  definition  of  a  circle,  it 
is  evident  that  all  its  radii  are  equal; 
and  also  that  all  its  diameters  are 
equal,  and  each  double  the  radius. 

180.  An  arc  of  a  circle  is  any  part  of  the  circumfer- 
ence, as  AED. 

A  semi'Circumference  is  an  arc  equal  to  one-half  the  cir- 
cumference. 

181.  A  chord  is  the  straight  line  which  joins  any  two 
points  on  the  circumference,*  as  AD. 

*  The  arc  is  sometimes  said  to  be  subtended  by  its  chord, 

74 


BOOK  1L-DEFINITI0N8.  75 

Chords  of  a  circle  are  said  to  be  equally  distant  from  the 
centre,  when  the  perpendiculars  drawn  to  them  from  the 
centre  are  equal.  The  chord  on  which  the  greater  perpen- 
dicular falls  is  said  to  he  further  from  the  centre, 

ROTE. — Every  chord  subtends  two  arcs  whose  sum  is  the  circumference.  A 
rd  which  does  not  pass  through  the  centre,  subtends  two  unequal  arcs. 
Thus,  AD  subtends  both  the  arc  AED  and  the  arc  ABCD;  of  these,  the  greater 
is  called  the  major  arc,  and  the  less  the  minor  arc.  Thus,  the  major  arc  is 
greater,  and  the  minor  are  less  than  the  semi-circumference. 

The  major  and  minor  arcs,  into  which  a  circumference  is  divided  by  a  chord, 
are  said  to  be  conjugate  to  each  other.  When  an  arc  and  its  chord  are  spoken 
of,  the  minor  arc  is  always  meant,  unless  otherwise  stated. 

182.  A  secant  of  a  circle  is  a  straight  line  of  indefinite 
length  which  cuts  the  circum- 
ference in  two  points,  as  AB. 

A  tangent  to  a  circle  is  any 
straight  line  which  meets  the 
circumference,  but,  being  pro- 
duced, does  not  cut  it,  as  CD. 
Such  a  line  is  said  to  touch  the 
circle  at  a  point,  and  the  point 
is  called  the  ji^om^  of  contact, 
or  point  of  tangency, 

A  secant  may  be  considered  as  a  chord  produced,  and  a  chord  may  be  con- 
sidered as  the  part  of  the  secant  that  is  within  the  circle. 

If  a  secant,  which  cuts  a  circle  at  the  points  P  and 
Q.  be  gradually  turned  about  P  as  fixed,  the  point 
Q  will  ultimately  approach  the  fixed  point  P.  When 
the  secant  PQB  reaches  this  limiting  position,  it  be- 
comes the  tangent  to  the  circle  at  the  point  P. 

183.  Two  circles  are  tangent  to 
each  other  when  they  are  tangent  to 
the  same  straight  line  at  the  same 
point. 

They  are  said  to  have  internal  con- 
tact or  external  contact,  according  as 
one  circle  is  entirely  within  or  en- 
tirely without  the  other. 


76 


PLANE  OEOMETRT. 


Circles  that  have  the  same  centre  are  said  to  be  con- 
centric. 

184.  A  segment  of  a  circle  is  the  figure 
included  between  a  chord  and  its  arc,  as 
ACB,  or  AFB. 

A  segment  is  called  a  major  or  minor 
segment,  according  as  its  arc  is  a  major  or 
mmor  arc.  The  chord  of  the  segment  is 
sometimes  called  the  base  of  the  segment. 

A  semicircle  is  the  segment  included  between  a  diameter 
and  a  semi-circumference,  as  DFE. 

185.  A  straight  line  is  inscribed  in  a 
circle  when  its  extremities  are  on  the  cir- 
cumference, as  AC. 

186.  An  inscribed  angle  is  one  whose 
vertex  is  in  the  circumference,  and  whose 
sides  are  chords  of  the  circle,  as  Z  ACB. 
The   /_  AGE,  whose  vertex  is  at  the  centre  0,  is  called  the 
angle  at  the  centre. 

187.  An  angle  in  a  segment  is  the  angle  formed  by  two 
straight  lines  drawn  from  any  point  in  the  arc  of  the  seg- 
ment to  the  extremities  of  the  base  of  the  segment,  as 
Z  ACB  or  ZADB. 

188.  A  sector  of  a  circle  is  the  figure 
bounded  by  two  radii  and  the  arc  inter- 
cepted between  them,  as  AOB,  or  BOC. 

189.  A  polygon  is  said  to  be  inscribed 
in  a  circle,  when  all  its  vertices  are  on  the 
circumference,  as  ABCD. 

190.  A  circle  is  said  to  be  circum- 
scribed about  a  polygon,  when  the  circum- 
ference passes  through  each  vertex  of  the 
polygon. 


BOOK  IL— DEFINITIONS.  Tl 


Vm  191.  A  polygon  is  said  to  be  circumscribed  about  a  circle , 
when  each  of  its  sides  is  tangent  to  the  circumference,  as 
ABCD.  The  circle  is  then  said  to  be  inscribed  in  the 
polygon. 


192.  From  the  definition  of  a  circle  it  follows; 

(1)  The  distance  of  a  point  from  the  centre  of  a  circle  is 
less  than,  equal  to,  or  greater  than  the  radius,  according 
as  the 2^oint  isicithin,  on,  or  without  the  circumference. 

Hence  (157),  the  locus  of  a  point,  which  is  always  at  a 
given  distance  from  a  given  point,  is  the  circumference  of  a 
circle,  of  which  the  given  point  is  the  centre,  and  the  given 
distance  is  the  radius, 

(2)  Circles  of  equal  radii  are  identically  equal.  For,  if 
one  circle  be  applied  to  the  other  so  that  their  centres  co- 
incide, their  circumferences  will  coincide,  since  all  the 
points  of  both  are  at  the  same  distance  from  the  centre 
(179). 

(3)  A  straight  line  cannot  meet  the  circumference  of  a 
circle  in  more  than  two  points.  For,  if  it  could  meet  it  in 
three  points,  these  three  points  would  be  equally  distant 
from  the  centre  (179).  There  would  then  be  three  equal 
straight  lines  drawn  from  the  same  point  to  the  same 
straight  line,  w^ich  is  impossible  (64). 


78  PLANE  GEOMETRY. 


Arcs  and  Chords. 


Proposition  1 .     Theorem. 

193.  Every  diameter  divides  the  circle  and  the  circum- 
ference into  two  equal  parts,  and  is  greater  than  any  other 
chord. 


Hyp,  Let  AB  be  the  diameter  of  the  0  ACB,  and  AC 
any  chord  not  passing  through  the  centre. 

To  prove  that  AB  bisects  the  O  and  the  Oce,  and  that 
AB  >  AC. 

Proof.  Join  OC.  Then  Jet  the  plane  containing  the 
arc  ACB  be  turned  about  AB  as  an  axis  until  it  falls  on 
the  plane  of  AEB. 

The  arc  ACB  will  coincide  with  the  arc  AEB,  since 
every  point  in  ACB  is  at  the  same  distance  from  the  centre 
0  as  every  point  in  AEB  (179). 

Hence  the  arc  ACB  =  the  arc  AEB, 

and  the  surface  ACB  =  the  surface  AEB.   (29) 

Also,  AC  <  AO  +  OC, 

i.e.,  AO  +  OC  >  AC. 

But  AO  +  OC  =  AO  +  OB, 

since  OC  and  OB  are  radii  of  the  same  ©  (179). 

,• .  AO  +  OB  >  AC,  or  AB  >  AC.         q.e.d. 


BOOK  II.—ARCS  AND   CHORDS,  79 


Proposition  2.    Theorem. 

194.  In  the  smne  circle,  or  in  equal  circles,  equal  arcs 
have  equal  chords,  and  suMend  equal  angles  at  the  centre, 

Hijp,    Let  ABO,  DEF  be  equal     ^ 
Os,  and  AB,  DE  equal  arcs. 

To  prove  I       C^      /  V       F 

chord  AB  =  chord  DE, 
and  Z  ACB  =  Z  DFE. 

Proof,  Apply  the  O  ABC  to  the  O  DEF,  so  that  the 
centre  C  may  fall  on  the  centre  F,  and  the  radius  CA  on 
the  radius  FD. 

Then,  because  the  Os  are  equal,  (Hyp.) 

.*.  the  pt.  A  will  fall  on  the  pt.  D, 

and  their  Oces  will  coincide, 

since  all  t/ie  pis.  of  both  are  at  iJie  same  distance  from  tJie  centre  (179). 
Because  arc  AB  =  arc  DE,  (Hyp.) 

.  • .  the  pt.  B  will  fall  on  the  pt.  E, 

and  the  chord  AB  will  coincide  with  chord  DE.   (Ax.  11) 
.-.  chord  AB  =  chord  DE. 

and  Z  ACB  =  Z  DFE.  Q.E.p. 

Note.— A  line,  straight  or  curved,  is  said  to  subtend  a  certain  angle  from  a 
certain  point  when  the  lines  drawn  from  the  point  to  the  ends  of  the  line  form 
that  angle.  Thus,  the  chord  AB  or  the  arc  AB  subtends  the  angle  ACB  frpm 
the  point  C. 

195.  Cor.  Sectors  of  eqtcal  arcs  in  the  same  or  equal 
circles  are  equal. 


80  PLANE  GEOMETRY. 


Proposition  3.    Theorem. 

196.  Li  the  same  circle,  or  in  equal  circles,  equal  angles 
at  the  centre  intercept  equal  chords  and  eqtial  arcs  on  the 
circumference. 

Hyp,  Let  ABO,  DEF  be  equal  A^m^  D^rTTr^E 
OS,  and  let  Z  C  =  Z  F.  (\/| 

To  prove  \      ^      /    V      ^ 

chord  AB  =  chord  DE,      V^_^ 

and  arc  AB  =  arc  DE. 

Proof.  Apply  the  O  ABO  to  the  O  DEF,  so  that  the 
centre  0  may  fall  on  the  centre  F,  and  the  radius  OA  on 
the  radius  FD. 

Then,  because  the  Os  are  equal,  (Hyp.) 

.*.  the  pt.  A  will  fall  on  the  pt.  D, 
and  their  ©ces  will  coincide.  (179) 

Because  Z  0  =  Z  F,  (Hyp.) 

.-.  OB  must  fall  on  FE; 
and  since  OB  =  FE,  (Hyp.) 

. ».  the  pt.  B  will  fall  on  the  pt.  E. 

!N"ow  since  A  coincides  with  D,  and  B  with  E,  and  the 
Oce  of  the  O  ABO  with  the  ©ce  of  the  O  DEF, 

.  • .  chord  AB  must  coincide  with  chord  DE,       (Ax.  11) 
and  the  arc  AB  must  coincide  with  the  arc  DE. 
.• .  chord  AB  =  chord  DE, 
and  arc  AB  =  arc  DE.  q.e.d. 

197.  Cor.  1.  Sectors  of  equal  angles  at  the  centre  in 
the  same  or  eqtial  circles  are  eqiiah 

198.  OoR.  2.  In  the  same  or  equal  circles  equal  chords 
subtend  equal  arcs  and  equal  ajigles  at  the  centre. 


BOOK  IL-^ARCS  AND  CHORDS.  81 


Proposition  4.    Theorem. 

I  199.  In  the  same  circle,  or  in  equal  circles,  of  two  un- 
equal minor  arcs,  the  greater  is  subtended  by  the  greater 
chord. 

D 


Hyp.     In  the  ©ABO,  let 

arc  AB  >  arc  AD. 
To  prove  chord  AB  >  chord  AD. 

Proof,      Draw  the  radii  CA,  CD,  CB. 
Because  the  arc  AB  >  the  arc  AD,  (Hyp.) 

.  • .  the  pt.  D  must  be  between  the  pts.  A  and  B, 
and  .-.  ZACB>ZACD. 

Then  in  the  A  s  ABC,  ADC,  we  have 

CB  =  CD,  CA  =  CA, 

ZACB>ZACD. 

.  • .  chord  AB  >  chord  AD. 
since  the  As  ham  two  sides  equal  each  to  eacJi,  and  the  included  As 
unequal  (119).  q.e.d. 

200.  Cor.  Conversely:  If  the  chord  AB  >  the  chord 
AD,  the  arc  AB  >  the  arc  AD.  For,  if  the  arc  AB  were 
equal  to  the  Jirc  AD,  the  chord  AB  would  be  equal  to  the 
chord  AD  (194);  and  if  the  arc  AB  were  less  than  the 
arc  AD,  the  chord  AB  would  be  less  than  the  chord  AD 
(199).  Til  en,  since  the  arc  AB  cannot  be  equal  to,  nor  less 
than  the  arc  AD,  it  must  be  greater. 

Hence,  the  greater  chord  subtends  the  greater  minor  arc* 


82 


PLANE  GEOMETRY. 


Proposition  5.    Theorem. 

201.   The  radius  which  is  perpendicular   to  a  chord 
bisects  the  chord  and/ the  arc  which  it  suUends, 

Hyp.    In  the  O  ABC  let  the  radius  OD 
be  _L  to  the  chord  AB  at  E. 
To  prove  AE  =  EB, 

and  arc  AD  =  arc  DB. 

Proof.  Join  OA,  OB. 

Then,  since  OA  =  OB,     (Radii) 

OE  is  common, 

rt.  zAEO^rt.ZBEO, 

.-.  aAOE=  aBOE. 


(Hyp.) 


Two  rt.  AS  are  equal  if  they  have  the  hypotenuse  and  a  side  equal 
each  to  each  (110). 


and 


.-.AE^^EB, 
ZAOE=  ZBOE. 


.•.  arc  AD  =  arc  DB, 

since  equal  ^sat  tJie  centre  intercept  equal  arcs  on  the  Qce  (196).  q.e.d. 

202.  Cor.  1.  TJie  perpendicidar  Usector  of  a  chord 
passes  through  the  centre  of  the  circle,  and  bisects  both  the 
arc  and  the  angle  at  the  centre  subtended  by  the  chord. 

203.  Cor.  2.  A  radius  which  bisects  a  chord  is  i^erpen- 
dicular  to  the  chord  and  bisects  the  subtended  arc. 

204.  Cor.  3.  A  radius  ichich  bisects  an  arc  is  perpen- 
dicular to  the  chord  of  that  arc  at  its  middle  point,  and 
bisects  the  angle  at  the  centre  which  the  arc  subtends. 

205.  Cor.  4.  The  locus  of  the  middle  point  of  parallel 
chords  in  a  circle  is  the  dia?mter  perpendicular  to  these 
chordsy 


BOOK  II.— ARCS  AND  CHORDS.  83 


Proposition  6.    Theorem. 

.       206.  In  the  smne  circle,  or  in  equal  circles ,  equal  chords 
(I  are  equally  distant  from  the  centre;  and,  conversely,  chords 
11    which  are  equally  distant  from  the  centre  are  equal  to  one 
I     another. 


I 


B 


(1)  Hyp.  Let  AB,  CD  be  equal  chords  in  the  O  ABDO, 
and  let  OF,OE  be  J.  to  AB,  CD. 

To  prove  OF  =  OE. 

Proof  Join  OA,  OC. 

Then,    since  OF,  OE  are  i.  to  AB,  CD, 

/.  AB,  CD  are  bisected  at  F,  E.  .  (201) 

But  AB  =  CD.  (Hyp.) 

.•.AP  =  CE.  (Ax.  7) 

Also,  AO  =  CO.  (Kadii) 

.-.  rt.  A  AOF  =  rt.  A  COE.  (110) 

.-.  OF  =  OE. 

(2)  Hyp.  Let  the  i.s  OF,  OE  be  equal. 

To  prove  AB  =  CD. 

Proof.     Since  OF  =  OE,  (Hyp.) 

and      *.  OA  =  OC,  (Eadii) 

.  •.  rt.  A  AOF  =  rt.  A  COE.  (110) 

.-.  AF  =  CE, 
and  .  • .  AB  =  CD.  (201  and  Ax.  6) 

Q.E.D. 

> 


84  PLANE  GEOMETRY. 

Proposition  7.    Theorem. 

207.  In  the  same  circle ,  or  in  equal  circles,  of  two  un- 
equal chords,  the  less  is  at  the  greater  distance  from  the 
centre. 

Hyj).  In  the  O  ABD,  let  the  chord 
CD  <  the  chord  AB,  and  let  the  J_s 
OE,  OF  be  drawn  from  the  centre  0 
to  CD,  AB. 

To  prove         OE  >  OF. 
Proof  I 

Since  chord  AB  >  chord  CD, 

.-.arc  AB  >arcCD.    (200) 
On  arc  AGB  take  arc  AG  =  arc  CD. 
Draw  the  chord  AG,  and  the  J_  OH. 

Since    arc  AG  =  arc  CD,  .  • .  chord  AG  =  chord  CD. 
Equal  arcs  have  equal  chords  (194). 

And  .-.OH^OE. 

Equal  chords  are  equally  distant  from  tJie  centre  (206). 

Because    arc  AG  <  arc  AB,  the  pt.  G  must  fall  within 

the  arc  AB. 

.* .  _L  OH  will  cut  the  chord  AB  at  some  pt.  K, 

Now,  OH  >  OK. 

The  whole  is  >  any  of  its  parts  (Ax.  8). 

And  OK  >  OF. 

T/ie  ±  is  the  shortest  distance  from  a  pt.  to  a  line  (58). 
.-.  a  fortiori,  OH  >  OF. 

.•.OE>OF.  Q.E.D 

208.  Cor.  Conversely,  of  two  chords  unequally  distant 
from,  the  centre,  the  one  which  is  at  the  greater  distance  is 
the  less, 

EXERCISE. 

If  two  chords  of  a  circle  cut  each  other,  and  make  equal 
angles  with  the  sti*aight  line  which  joins  their  point  of 
iutersection  to  the  centre,  prove  that  the  chords  are  equal., 


BOOK  II.—THE  CIHGLE.     TANGENTS,  85 


Proposition  8.    Tiieorem. 

7       209.  A  straight  line  'perpendicular  to  a  radius  at  its 
extremity  is  a  tangent  to  the  circle. 


Hyp.    Let  0  be  the  centre  of  a  O,  OA  the  radius,  and 
BC  a  line  J.  to  OA  at  A. 

To  prove    BC  tangent  to  the  O. 

Proof,  In  BC  take  any  pt.  D,  other  than  A;  join  OD. 

Then,  since  OA  is  _L  to  BC,  (Hyp.) 

.-.  OA  <  OD. 

Tlie  L  is  the  shortest  distance  from  a  pt.  to  a  line  (58). 


and 


.'.  the  pt.  D  is  without  the  circle.       [192.  (1)] 
BC  has  every  pt.  except  A  without  the  O, 

.  • .  BC  is  a  tangent  to  the  O  at  A.  (182) 

Q.E.D. 


210.  Cor.  1.  Conversely,  a  tangent  to  a  circle  at  any 
point  is  perpendicular  to  the  radius  drawn  to  that  point, 

211.  Cor.  2.  The  perpendicular  to  a  ta^igent  at  the  point 
of  tangency  passes  though  the  centre  of  the  circle, 

212.  Cor.  3.  The  straight  line  drawn  from  the  centre 
perpendicular  to  the  tangent  meets  it  in  the  point  of  con- 
tact, 

213.  Cor.  4.  Ofily  one  tangent  can  he  drawn  to  a  circle 
at  a  given  point  on  the  circumference. 


86 


PLANE  GEOMETRY. 


i 


Proposition  9.    Theorem. 

214.  Two  parallel  lines  intercept  equal  arcs  on  the  cir- 
cumference. 

There  may  he  three  cases. 
Hyp.    Let  AB,  CD  be  the  ||  s. 
Case  I.   When  AB,  CD  are  secants. 
To  prove    arc  AC  =  arc  BD. 
Proof.  Draw  the  radius  OE  J_  to  one 
of  the  II  s. 
It  will  then  be  J.  to  the  other  || . 

A  St.  line  ±  to  one  of  two  \\sis  ±  to  the  other  (71). 

.-.  arc  AE  =  arc  BE,  and  arc  CE  =  arc  DE,  (201) 

.  • .  arc  AC  =  arc  BD.  (Ax.  3) 

Case  II.   When  AB  is  a  tangent  and  CD  is  a  secant. 
To  prove   arc  CE  =  arc  DE. 
Proof.  Draw  the  radius  OE  to  the  pt. 
of  contact  E. 
Then  OE  is  J.  to  AB     (210), 

and  to  its  ||  CD       (71). 
.-.  arcCE  =  arcDE,       (201). 
Case  III.    When  AB,  CD  are  tangents. 
To  prove 

arc  EMH  =  arc  ENH. 
Proof.  Draw  the  secant  MN  ||  to  AB. 
Then        arc  ME  =  arc  NE, 
and         arc  MH  =  arc  NH. 
Adding, 

arc  EMH  =  arc  ENH.  q.e.d. 

215.  Cor.  1.  Conversely,  if  the  arcs  intercepted  hy  tico 
secants  are  equal,  the  secants  are  parallel. 

216.  Cor.  2.   Tlte  straight  line  joining  the  points  of  coiir 
tact  of  tioo  parallel  taiigents  is  a  diameter. 


l]^^ 


ase  II) 


¥ 


BOOK  II.-THE  CIRCLE.  87 

Proposition  lO.    Thieorem. 
217.  Tliroxigh  three  given  points  not  in  the  saine  straight 
line,  one  circumference,  and  only  one,  can  be  draion, 

I  ^^^^.  ;^      I 
\  "\    I?/ 

B 

Hyp,  Let  A,  B^  0  bo  the  three  given  pts.  not  in  a  st. 
line. 

To  prove  that  one  Oce,  and  only  one,  can  be  drawn 
through  A,  B,  0. 

Proof,  Join  AB,  BO. 

Bisect  AB,  BC  by  the  _Ls  DF,  EG. 

Since       AB,  BC  are  not  in  the  same  st.  line,        (Hyp-) 
.*.  the  JLs  DF,  EG  must  meet  at  some  pt.  0.        (76) 

Because  G  is  in  the  J_  DF, 

.'.  it  is  equidistant  from  A  and  B.  (66) 

And  because  G  is  in  the  J_  EG, 

.  *.  it  is  equidistant  from  B  and  0.  (66) 

Then,      because  G  is  equidistant  from  A,  B,  C, 

.-.  the  Oce  described  with  centre  G  and  radius  GA  will 
pass  through  A,  B,  C. 

Again,  only  one  Oce  can  be  so  described. 

For  if  any  Oce  pass  through  A,  B,  C,  its  centre  will  be 
at  once  in  the  J_  bisectors  DF,  EG,  and  .*.  at  their  pt.  of 
intersection.  But  two  st.  lines  cannot  intersect  in  more 
than  one  pt. 

.*.  there  is  only  one  Oce  that  can  pass  through  A,  B,  and  C. 

Q.E.D. 

218.  Cor.  1.  Tivo  circumferences  cannot  intersect  in 
more  than  two  points, 

219.  Cor.  2.  Two  circumferences  ivhicli  have  three 
points  cojnmon  coincide. 


PLANE  GEOMETRY. 


Relative  Position  of  Two  Circles. 


Proposition  1  1 .    Theorem. 

220.  If  Uoo  circumfeveiices  intersect  each  other,  the  right 
line  joining  their  centres  bisects  their  common  chord  at 
right  angles. 

Hyp,  Let  0,  0'  be  the  centres 
of  two  Oces  which  intersect  each 
other;  and  A,  B  their  pts.  of  inter- 
section. 

To  2Jrove  that  the  line  00'  bi- 
sects AB  at  rt.  Z  s. 

Proof.  Because  0  and  0'  are  each  equally  distant  from 
AandB,  (179) 


.•.  the  line  00'  bisects  AB  at  rt  Z  s. 


(67) 


221.  CoR.  1.  Conversely,  the  perpendicular  bisector  of 
a  common  chord  passes  through  the  centres  of  both  circles, 

222.  Cor.  2.  If  we  suppose  the 
circles  to  be  moved  so  that  the 
point  A  approaches  the  line  00', 
tlie  pt.  B  will  also  approach  the 
line;  and  since  the  line  00'  is 
always  perpendicular  to  the  mid- 
dle of  AB  (220),,  the  two  points 
A  and  B  will  ultimately  come  together  on  the  line  00', 
and  be  united  in  a  single  point  common  to  the  two  circles. 
The  common  chord  AB  will  then  be  a  common  tangent  to 
the  two  circumferences  at  their  point  of  contact. 

Hence,  when  tivo  circumferences  are  tangent  to  each 
other,  their  point  of  contact  is  in  the  straight  line  joining 
their  centres ;  and  the  perpe7idicular  at  this  2)oint  is  a 
common  tangent  to  the  two  circumferences. 


1 


BOOK  II.—TWO  CIRCLES. 


Proposition  12.    Theorem. 

223.  If  two  circumferences  intersect  each  other,  the  dis- 
tance between  their  centres  is  less  than  the  sum  and  greater 
Vian  the  difference  of  the  radii. 


\ 


Hyp.  Let  the  Os  with  centres  0,  0'  intersect  at  A. 
Join  00',  AO,  AO'. 

To  yrove  00'  <  OA  +  AO',  and  >  OA  -  AO'. 
Proof.     In  the  aOAO' 

00'  <  OA  +  AO',  and  00'  >  OA  -  AO'. 
Eiilier  side  of  a  t^  <  the  sum  and  >  ilie  difference  of  the  other  two 
sides  (96).  q.e.d. 

224.  Cor.  1.  If  tlie  distance  of  the  centres  of  two  circles 
is  greater  than  the  sum  of  their  radii,  they  are  tvholly 
exterior  to  each  other. 

2t2ib.  Cor.  2.  If  the  distance  of  the  centres  of  two  circles 
is  equal  to  the  sum  of  the  radii,  they  are  tangent  exter- 
nallij. 

226.  Cor.  3.  If  the  distance  of  the  centres  is  less  than 
the  sum  and  greater  than  the  difference  of  the  radii,  the 
circles  intersect. 

227.  Cor.  4.  If  the  distarice  of  the  centres  is  equal  to 
the  diff^erence  of  the  radii,  the  circles  are  tangent  internally. 

228.  Cor.  5.  If  the  distance  of  the  centres  is  less  than 
the  diff'erencel^of  the  i^adii,  one  circle  is  wholly  luithin  the 
other. 

ScH.  If  two  circles  intersect  and  the  radius  of  either 
circle  drawn  to  a  point  of  section  touches  the  other  circle, 
the  circles  intersect  orthogonally,  i.e.,  at  right  angles. 


90  PLANE  GEOMETRY. 

EXERCISES. 

1.  Through  a  given  point  P  either  Jnside  or  outside  a 
given  circle  whose  centre  is  0,  two  straight  lines  PAB, 
PCD  are  drawn  making  equal  angles  with  OP,  and  cut- 
ting the  circle  in  A,  B,  C,  D:  prove  that  AB  =  CD,  and 
PA  =  PC. 

2.  P  is  a  point  inside  a  circle  whose  centre  is  0 :  prove 
that  the  chord  which  is  at  right  angles  to  OP  is  the  shortest 
chord  that  can  be  drawn  through  P. 

Let  APB  be  ±  to  OP,  CPD  any  other  chord  through  P:  draw  OE  ±  to  CD,  etc. 

3.  If  two  circles  cut  each  other,  any  two  parallel  straight 
lines  drawn  through  the  points  of  intersection  to  cut  the 
circles  are  equal. 

4.  Two  circles  whose  centres  are  A  and  B  intersect  at  C; 
through  C  two  chords  DCE,  FCG  are  drawn  equally  in- 
clined to  AB  and  terminated  by  the  circles:  show  that 
DE  =  FG. 

5.  Prove  that  the  two  tangents  drawn  to  a  circle  from  an 
external  point  are  equal  and  equally  inclined  to  the  straight 
line  joining  the  point  to  the  centre  of  the  circle. 

6.  A  is  a  point  outside  a  given  circle  whose  centre  is  0; 
with  centre  A  and  radius  AO  a  circle  is  described,  and 
with  centre  0  and  radius  equal  to  the  diameter  of  the  given 
circle  another  circle  is  discribed  cutting  the  last  in  B;  OB 
is  joined  cutting  the  given  circle  in  E:  prove  that  AE  is 
tangent  to  the  given  circle. 

The  Measukement  of  Angles. 

229.  To  measure  a  quantity  is  to  find  how  many  times 
it  contains  another  quantity  of  the  same  kind  taken  as  a 
standard  of  comparison.     This  standard  is  called  the  irnit. 

Thus,  if  we  wish  to  measure  a  line,  we  must  take  a  unit 
of  lengtlif  and  see  how  many  times  it  is  contained  in  the 
line  to  be  measured. 


BOOK  IL—TIIE  CIRCLE.  91 

The  number  which  shows  how  many  times  a  quantity 
contains  the  unit,  is  called  the  numerical  measure  of  that 
quantity. 

230.  The  rehitive  magnitude  of  two  quantities,  meas- 
ured by  the  number  of  times  which  the  first  contains  the 
second,  is  called  their  ratio.    Thus,  the  ratio  of  A  to  B  is 

y  ,  or  A  :  B. 

Since  the  ratio  of  two  quantities  is  found  by  dividing  the 
first  by  the  second,  therefore  the  ratio  of  two  quantities  is 
the  number  which  would  express  the  measure  of  the  first, 
if  the  second  were  taken  as  unity. 

The  ratio  of  two  quantities  is  the  same  as  the  ratio  of 
their  numerical  measures. 

Thus,  if  A  contains  the  unit  rn  28  times,  and  B  contains 

.,  ^  ^.  ,         A       28w       28 

it  9  times,  we  have  ^-  =  -7: —  =  -7^ . 
B        9m         9 

231.  When  a  quantity  is  contained  an  exact  number  of 
times  in  two  quantities  of  its  kind,  it  is  called  their  com- 
mon  measure. 

Two  quantities  are  commensurable  when  they  have  a 
common  measure. 

The  ratio  of  two  commensurable  quantities  can  be  ex- 
pressed by  a  whole  number  or  by  a  fraction. 

Thus,  if  each  of  the  two  lines  A  and  B  contains  some 

line  0  an   exact  number  of  times.   A' ^ ^ ■■ • 

as  for  example  if  A  contains  it  5  ^^___^___^__^___^ 
times  and  B  contains  it  4  times, 

the  two  lines  A  and  B  are  com-  C' • 

mensurable,  and   the   line  C   is  their   common  measure. 
Their  ratio  is  expressed  by  the  fraction  f . 

If  0  is  not  contained  an  exact  number  of  times  in  A  and 
B,  but  if  there  be  a  common  measure  which  is  contained, 
say,  25  times  in  A  and  19  times  in  B,  then  the  ratio  of  A 
to  B  is  the  fraction  f f . 


92  PLANE  GEOMETRY. 

Generally,  if  there  be  a  common  measure  which  is  con- 
tained m  times  in  A  and  7i  times  in  B,  their  ratio  will  be 
m 
n  ' 

232.  Two  quantities  are   incommensuralle  when  they 

have  no  common  measure.     The  ratio  of  such  quantities  is 

^   ^^jcalM  an  incornmcnsurable  ratio.     This  ratio  cannot  be  ex- 

^^^^actlyexpressed  in  figures;  but  its  numerical  value  can  be 

obtained  approximately  as  near  as  we  please. 

Thus,  suppose  A  and  B  are  two  lines  whose  ratio  is  V2, 
We  cannot  find  any  fraction  which  is  exactly  equal  to 
V"2',  but  by  taking  a  sufficient  number  of  decimals,  we 
may  find  V^  to  any  required  degree  of  approximation. 

Thus,  V^=  1.4142135  .  .  .  .  , 

and  therefore       1^2  >  1.414213     and     <  1.414214. 

That  is,  the  ratio  of  A  to  B  lies  between  iHo f  o  o  ^^^ 
To-o MM  y  ^^^d  therefore  differs  from  either  of  these  ratios 
by  less  than  one-millionth.  And  since  the  decimals  may 
be  continued  without  end  in  extracting  the  square  root  of 
2,  it  is  evident  that  this  ratio  can  be  expressed  as  a  fraction 
with  an  error  less  than  any  assignable  quantity. 

In  general,  when  A  and  B  are  incommensurable,  divide 
B  into  n  equal  parts  each  equal  to  x,  so  that  B  =  nx,  where 
n  is  an  integer.     Also  let  A  >  mx  but  <(y;i  +  1)^;;  tlien 

A  ^  mx         T       ^  Un  +  l)x 

^  >  —     and     <  ^^ ' — '-  : 

B       nx  nx 

that  is,  :jT  lies  between  —   and  -;  so  that  ^  differs 

B  n  n  n 

m  .  1 

from  —  by  a  quantity  less  than  — .     And  since  n  can  be 

taken  as  great  as  we  please,  -  may  be  made  as  small  as  we 

n 


BOOK  IL—MEASUnEMENT  OF  ANGLES.  93 

please  until  it   becomes  less  than  any  assignable  value, 
though  it  can  never  reach  absolute  zero. 

We  say,  therefore,  that  zero  is  the  limit  of  -  as  n  is  in- 
creased indefinitely. 

Hence,  Uvo  integers  can  he  found  whose  ratio  loill  express 
the  ratio  of  two  incominensurable  quantities  to  any  re- 
quired degree  of  accuracy. 

233.  Theorem.  Two  incommensuraUe  ratios  are  equal, 
if  their  approximate  numerical  values  always  remain  equal 
while  the  common  measure  is  iiidefi^iitely  diminished. 

Hyp,  Let  A:B  and  A':  B'  be  two  incommensurable 

m 
ratios,  whose  true  values  always  differ  from   —   by  less 

than  — . 
n 


To  prove         '        A:B  =  A':B'. 

Proof  Since  each  of  the  ratios  ^,  :^,  differs  from  -  by 
less  than  — , 


A         A'  1 

the  difference  between  ^r  and  vr>  must  be  <  - . 
B  B'  n 


But,  by  diminishing  the  common  measure,  n  can  be  made 

as  great  as  we  please,  and  therefore  —  may  be  made  as  small 

as  we  please,  i,  e.,  less  than  any  assignable  quantity,  however 
small.  Hence,  the  difference  between  A  :  B  and  A':  B' 
must  be  less  than  any  assignable  quantity,  however  small. 

.-.  A:B  =  A':B'. 


94  PLANE  GEOMETRY. 

Proposition  13.    Theorem. 

234.  In  the  same  circle,  or  in  equal  circles,  angles  at  the 
centre  are  in  the  same  ratio  as  their  intercepted  arcs. 

Hyp,  Let  AOB,  AOC  be  any  two 
Z  s  at  the  centres  of  two  equal  Os,      /^  ^\      /^^^^\ 
and    AB,    AC    their    intercepted    [        O       \(       o        j 

arcs.  \  /  ' '  ''/V/  /  'i  \/ •  '  ^*  ^\/d 

ZAOB      arcAB       ^^-i-^Ll /nLL>^^ 
To  prove         ^^^  = -_^_.  „ 

Case  I.  When  the  arcs  are  commenstiraUe, 

Proof.  Take  M,  any  common  measure  of  AB  and  AC, 

and  suppose  it  to  be  contained  five  times  in  AB  and  four 

times  in  AC. 

rrn  arc  AB      5 

Then  ^-^  =z  -  (1) 

arc  AC      4  ^  ^ 

Draw  radii  to  tlie  several  pts.  of  division  of  tlie  arcs  AB, 
AC,  dividing  the  Z  AOB  into  five  /  s  and  /  AOC  into 
four  Zs. 

These  Z  s  are  all  equal. 
In  the  same  ©  or  in  equal  Qs  equal  arcs  subtend  equal  /.s  at  the 
ceirtre  (194). 
ZAOB      5 


(2) 
(Ax.  1) 


'  *   ZA0C""4' 
Therefore,  from  (1)  and  (2), 

ZAOB  _  arc  AB 
Z  AOC  "arc  AC' 

Case  II.   When  the  arcs  are  incomme^isurahle. 

Proof.  In  this  case  we  know 
(232)  that  we  may  always  find  an 
arc  AD  as  nearly  equal  as  2veplease 
to  AC,  and  such  that  AB,  AD  are 
commensurable. 

Join  OD;  then 

Z  AOB      arc  AB  ,^       ^, 

"^XrvTk  = tt^'  (Case  J ) 

Z  AOD      arc  AD  '^  ^ 


BOOK  n.-MEASUREMENT  OF  ANGLES.  95 

Now,  these  two  ratios  being  always  equal  while  the  com- 
mon measure  is  indefinitely  diminished,  they  will  be  equal 
when  D  moves  up  to  and  as  nearly  as  we  please  coincides 
with  0.  (233) 


I 


Z  AOB      arc  AB 

ZAOG~arc  AG*  ^ 


235.  CoE.  Li  the  same  circle,  or  in  equal  circles,  sectors 
are  in  the  same  ratio  as  their  arcs;  for  sectors  are  equal 
when  their  arcs  are  equal.  (195) 

236.  ScH.  Since  the  angle  at  the  centre  of  a  circle,  and 
the  arc  intercepted  by  its  sides  increase  and  decrease  in 
the  same  ratio  (234),  the  numerical  measure  of  the  angle 
is  the  same  as  that  of  the  arc.  This  theorem,  being  of  very 
frequent  use,  is  expressed  briefly  by  saying  that  an  angle  at 
the  centre  is  measured  hy  its  interce'pted  arc.^  This  means 
simply  that  an  angle  at  the  centre  is  the  same  part  of  the 
whole  angular  magnitude  about  the  centre  that  its  inter- 
cepted arc  is  of  the  whole  circumference. 

237.  The  circumference  is  divided,  like  the  angular 
magnitude  about  the  centre  (28),  into  360  equal  parts  called 
degrees.  The  degree  is  divided  into  60  equal  parts  called 
minutes,  and  the  minute  into  60  equal  parts  called  seconds. 
Hence  the  unit  of  angle  and  the  unit  of  arc  are  both  called 
a  degree.  When  the  angle  becomes  a  right  angle  (20),  the 
arc  becomes  a  quarter  of  the  circumference,  or  a  q\iadrant. 
When  the  angle  becomes  a  straight  angle  (21),  the  arc  be- 
comes a  semi-circumference,  and  so  on.  A  right  angle  and 
a  quadrant  are  both  expressed  by  90".  Two  right  angles 
and  a  semi-circumference  are  both  expressed  by  180°.  Four 
right  angles  and  a  circumference  are  both  expressed  by  360°. 

*Boucb6  et  Comberousse,  p.  64. 


96 


PLANE  GEOMETUY. 


I 


Proposition  14.    Theorem. 

238.  Ail  inscribed  angle  is  measured  hij  one-half  the  arc 
intercepted  between  its  sides. 

Hyp.    Let  BAG  be  an  Z  inscribed  in 
tlie  OABO  and  intercepting  the  arc  BO. 

To  prove  Z  BAG  is  measured  by  ^  arc  BG. 

Gase  I.    When  tlie  centre  0  is  loithin 
the  /BAG. 

Proof.  Join  AO,  produce  it  to  D,  and 
join  OB,  OG. 

In  aAOB, 


Z  BOD  =  Z  OAB  +  Z  OBA. 
The  ext.  /.  of  a  A  equals  the  sum  of  the  opp.  int.  Z«  (98). 


But,  since  OA 


OB, 

.-.  Z  OAB  =  Z  OBA, 

being  opp.  equal  sides  (111). 

.-.  Z  BOD  =  21  OAB. 

Similarly,    Z GOD  =  2Z OAG. 

.-.whole  ZBOG  :=2zBAG. 

But  Z  BOG  is  measured  by  arc  BC. 

The  Z  at  the  centre  is  measured  by  the  intercepted  arc  (236) 

.*.  2  Z  BAG  is  measured  by  arc  BO. 

.*.     Z  BAG  is  measured  by  ^  arc  BG. 

Gase  II.   Wlien  the  centre  0  is  ivithoiit 
the  ZBAG. 

Proof.  Join  AO,  produce  it  to  D,  and 
join  OB,  OG.     Then, 

Z  DOB  =  2  Z  DAB,  (Gase  I) 
and  Z  DOG  =  2  Z  DAG.  (Gase  I) 

.•.ZD0G-ZD0B  =  2ZDAG  -2ZDAB. 
.-.Z  BOG  ==2  ZBAG. 


(Radii) 


(Ax.  2) 


f 


BOOK  TI.~MEASUREMENT  OF  ANQLE8.  97 

But  Z  BOC  is  measured  by  arc  EC.  (236) 

.-.  Z  BAG  is  measured  by  ^  arc  BC.  q.e.d. 

Note.— This  theorem  is  equally  true  when  the  angle  at  the  centre  is  greater 
than  two  right  angles,  as  the  student  may  show. 


239.  Cor.  1.  All  migles  inscribed    q 
in  the  same  segment  are  equal;   for 
each  is  measured  by  one-half  the  same 
arc  AFB. 


240.   Cor.   2.  Every  angle  AHB, 
inscribed  in  a  semicircle,  is  a  right 
angle;  for  it  is  measured  by  one-half 
a  semi-circumference,  or  by  a  quad-  A 
rant  (237). 


241.  Cor.  3.  Evertj  angle  BAC, 
inscribed  in  a  segment  greater  than  a 
semicircle,  is  an  acute  angle;  for  it  is 
measured  by  one-half  the  arc  BDC, 
which  is  less  than  a  quadrant. 


Every  a7igle  BDC,  inscribed  in  a  segment  less  than  a 
semicircle,  is  an  obtuse  aiigle;  for  it  is  measured  by  one- 
half  the  arc  BAC,  which  is  greater  than  a  quadrant. 

242.  Cor.  4.  The  opposite  angles  of  an  inscribed  quad- 
rilateral are  supplementary ;  for  the  sum  of  the  Zs  A  and  ,._-- 
D  isjneasured  by  one-half_the  Qce,  which  is  the  measure    '^ 
^ritwo  right  Z s  (237);  therefore  the  Zs  are  supplement- 
ary (25). 


§B  PLANE  GEOMETRY. 

Proposition    1 5.    Tiieorem. 

243.  An  angle  formed  ly  a  tangent  and  a  chord  from  the 
point  of  contact  is  measured  iy  one-half  the  intercepted  arc. 

Hyp,  Let  AC  be  a  tangent  to  the  ^ 

OBHE  at  B,  and  BD  any  chord  of  the  ^-^ 

Ofrom  B.  / 

To  prove      Z  CBD  is  measured  by     / 
i  arc  BHD.  ( 

Proof.  From  B  draw  BE  at  rt.  Z  s     \ 
to  AC.'  \^^ 

BE  is  a  diameter  of  the  O.  ^  ^  C 

TJie  1  to  a  tangent  at  the  pt.  of  contact  passes  tJiroiigh  the  centre 
of  the  O  (311). 
Then,  rt.   Z  CBE  is  measured  by  J  semiOce  BHE, 
and  Z  DBE  is  measured  by  \  arc  DE.  (238) 

Therefore,  subtracting, 

Z  CBD  is  measured  by  \  arc  BHD. 
Similarly,  Z  ABD  is  measured  by  ^  arc  BED.  Q.E.D. 

244.  ScH.  This  proposition  is  a  particular  case  of  Prop. 
14.  Thus,  let  the  side  BD  remain  fixed,  while  the  side  BH 
turns  about  B,  as  in  (28),  until  it  becomes  the  tangent  BC 
at  the  point  B.  In  every  position  of  the  chord  BH,  the 
inscribed  angle  HBD  is  measured  by  half  the  intercepted 
arc  HD.  Therefore,  when  the  chord  BH  becomes  the  tan- 
gent BC,  the  angle  CBH  is  measured  by  half  the  arc  BHD. 

EXERCISES. 

1.  If  the  angle  BAC  at  the  circumference  of  a  circle  be 
half  that  of  an  equilateral  triangle,  prove  that  BC  is  equal 
to  the  radius  of  the  circle. 

2.  If  a  hexagon  be  inscribed  in  a  circle,  show  that  the 
sum  of  any  three  alternate  angles  is  four  right  angles. 

3.  If  two  circles  intersect  in  the  points  A,  B,  and  any 
two  lines  ACD,  BFE,  be  drawn  through  A  and  B,  cutting 
one  of  the  circles  in  the  points  C,  E,  and  the  other  in  the 
points  D,  F,  the  line  CE  is  parallel  to  DF. 


BOOK  II.-MEASUREMENT  OF  ANGLES. 


99 


Proposition  1 6.    Theorem. 

245.  An  angle  formed  ly  two  cJiords  wliich  intersect 
witlmi  a  circlcy  is  measured  by  one-half  the  sum  of  the  arcs 
intercepted  hetweenits  sides  and  between  its  sides  2)roduced. 


Hyp,  Let  BD,  CE  be  two  chords  intersecting  at  A 
within  the  O  BODE. 

To  prove  Z  BAG  is  measured  by  ^  (arc  BO  +  ^rc  DE). 
Proof  Join  BE. 

ZBAC  =  ZAEB  +  ZABE. 
TJie  ext.  /_  of  a  t.  equals  tJie  sum  of  the  opp.  int.  Zs  (98). 

But  Z  AEB  is  measured  by  J  arc  BO, 

and  Z  ABE  is  measured  by  J  arc  DE.  (238) 

adding,    Z  BAO  is  measured  by  J  (arc  BO  +  arc  DE). 

Q.E.D. 

EXERCISES. 

1.  If  arc  BO  =  84°  and  Z  CAD  is  a  rt.  I,  how  many  de- 
grees are  there  in  the  arc-DE  ? 

2.  The  side^  of  a  quadrilateral  touch  a  circle,  and  the 
straight  lines  drawn  from  the  centre  of  the  circle  to  the 
vertices  cut  the  circumference  in  A,  B,  0,  D :  show  that 
AC,  BD,  which  intersect  inside  the  circle,  are  at  right 
angles  to  each  other. 


100 


PLANE  GEOMETRY. 


Proposition  1  7.    Theorem. 

246.  A7i  angle  formed  hy  Uvo  secants  which  intersect 
loithout  a  circle,  is  measured  by  one-half  the  difference  of 
the  intercepted  arcs. 

Hyp.    Let  AC,  AB  be  two  secants 
intersecting  at  A  without  the  O  BCED. 
To  prove  Z  A  is  measured   by 

\  (arc  BC  -  arc  DE). 
Proof,  Join  BE. 

ZBEO=ZA+ZB  (98) 

.•.ZA=  ZBEC-  ZB. 
But   Z  BEC  is  measured  by  ^  arc  BC. 
and   Z  B  is  measured  by  \  arc  DE.   (238) 

.*.  Z  A  is  measured  by  |  (arc  BC  —  arc  DE).        q.e.d. 

247.  ScH.  Prop.  14  may  be  considered  as  a  special  case 
of  Props.  J  6  and  17  by  conceiving  AB  in  (245)  and  (246)  to 
move  parallel  to  its  present  position  until  D  reaches  E. 
When  D  reaches  E,  the  arc  DE  becomes  zero,  and  BAC  be- 
comes an  inscribed  angle,  measured  by  half  its  intercepted 
arc. 


EXERCISES. 


1.  If  arcBC  =  80°  and  zB 
degrees  in  the  angle  A.    /  , 


14°,  find  the  number  of 


w  2.  A,  B,  C  are  three  points  on  the  circumference  of  a 
circle,  the  bisectors  of  the  angles  A,  B,  C  meet  in  D,  and 
AD  produced  meets  the  circle  in  E :  prove  that  ED  =  EC. 
3.  If  a  quadrilateral  be  described  about  a  circle,  the 
angles  at  the  centre  subtended  by  the  opposite  sides  are 
supplemental. 


BOOK  IL-MEASUliEMENT  OF  ANGLES,  101 


Proposition  18.    Theorem. 

248.  An  angle  formed  hy  a  tangent  and  a  secant  is 
measured  hy  one-half  the  difference  of  the  intercepted  arcs. 


C 

H 

Hy]?.  Let  AC,  AB  be  a  tangent  and  a  secant  intersect- 
ing at  A. 

To  2Jrove  Z  A  is  measured  by  \  (arc  BHE  —  arc  DE). 
Proof    Join  BE. 

ZBEC  =  ZA+ZB  (98) 

.-.   ZA=:  ZBEO-ZB. 
But     Z  BEG  is  measured  by  |  arc  BHE,  (238) 

and  Z  B  is  measured  by  ^  arc  DE.  (238) 

.'.   Z  A  is  measured  by  \  (arc  BHE  —  arc  DE). 

Q.E.D. 

249.  Cor.  The  angle  formed  hy  ttoo  tangents  is  meas- 
ured hy  one-half  the  difference  of  the  intercepted  arcs, 

EXERCISES. 

1.  Two  tangents  AB,  AC  are  drawn  to  a  circle;  D  is  any 
point  on  the  circumference  outside  the  triangle  ABC:  show 
that  the  sum  of  the  angles  ABD  and  ACD  is  constant. 

2.  If  a  variable  tangent  meets  two  parallel  tangents  it 
subtends  a  right  angle  at  the  centre. 


102  PLANE  GEOMETRY. 

Quadrilaterals. 
Proposition    1 9.    Theorem. 

250.  If  the  opposite  angles  of  a  quadrilateral  are  sup- 
plementary, the  quadrilateral  may  he  inscribed  in  a  circle. 

Hyp.  Let  ABCD  be  a  quadrilateral  in 

wliich         ZB+  ZD=2rt.  /s. 

To  prove  the  pts.  A,  B,  C,  D  are  in 
the  same  O. 

Proof  Through  the  three  pts.  A,  B, 
C  describe  a  O. 

If  this  O  does  not  pass  through  D,  it  "" '" 

will  cut  AD,  or  AD  produced,  at  some  other  pt.  than  D. 

Let  E  be  this  pt.     Join  EC. 

Because  the  quadrilateral  ABCE  is  inscribed  in  a  O, 

.-.   ZABC+ zAEC  =  2rt.  Zs. 
T?ie  opp.  Z.sof  an  inscribed  quad,  are  supplementary  (242). 

But         Z  ABC  +  Z  ADC  =  2  rt.  Z s.  (Hyp.) 

.-.   Z  ABC  +  Z  AEC  =  Z  ABC  +  Z  ADC.     (Ax.  1) 
.-.   ZAEC=  ZADC;  (Ax.  3) 

that  is,  an  ext.  Z  of  a  A  ==  an  int.  opp.  Z ,  which  is  im- 
possible. (98) 
.*.    the  circle  which  passes  through  A,  B,  C,  must  pass 
through  D.                                                                       q.e.d. 

251.  Def.  Points  which  lie  on  the  circumference  of  a 
circle  are  called  concyclie. 

A  cyclic  quadrilateral  is  one  which  is  inscribed  in  a 
circle. 

EXERCISE. 

If  two  opposite  sides  of  a  cyclic  quadrilateral  be  pro- 
duced to  meet,  and  a  perpendicular  be  let  fall  on  the  bi- 
sector of  the  angle  between  them  from  the  point  of  inter- 
section of  the  diagonals :  prove  that  this  perpendicular  will 
bisect  the  angle  between  the  diagonals, 


BOOK  IL—qUADlULATEHALS, 


103 


V 


Proposition  20.    Theorem. 

252.  hi  any  quadrilateral  circiunscribing  a  circle,  the 
um  of  one  pair  of  opposite  sides  is  equal  to  the  sum  of  the 
ther  pair. 


Hyp,  Let  ABCD  be  a  quadri- 
lateral circumscribing  a  O. 

To  prove  AB  +  CD  =  AD  +  BO. 

Proof.  From  the  centre  0  draw 
the  radii  to  the  pts.  of  contact  E, 
Y,  G,  H,  and  draw  OB. 

Thenrt.AOBE=rt.AOBF,(no)    ^ 

.-.  EB  =  rB. 


Similarly,    EA  =  HA,  GD  =  HD,  GO  =  FO. 
Adding  these  four  equations,  we  have 

EB  +  EA  +  GD  +  GC  =  FB  +  HA  +  HD  +  FO, 
or         AB  +  CD  =  BC  +  AD.       q.e.d. 


EXERCISES. 

1.  The  Ime  joining  the  middle  points  of  two  parallel 
chords  of  a  circle  passes  through  the  centre. 

2.  The  chords  that  join  the  extremities  of  two  equal 
arcs  of  a  circle  towards  the  same  parts  are  parallel. 

3.  The  sum  of  the  angles  _subtended  at  the  centre  of  a 
circle  by  two  opposite  sides  of  a  circumscribed  quadrilateral 
is  equal  to  two  right  angles. 


104  PLANE  GEOMETRY. 

4.  An  isosceles  triangle  has  its  vertical  angle  equal  to  the 

exterior  angle  of  an  equilateral  triangle.     Prove  that  the 

radius  of  the  circumscribing  circle  is  equal  to  one  of  the 

equal  sides  of  the  given  triangle. 

.       5.  Prove  that  the  radius  of  the  circle  inscribed  in  an 

^(.equilateral  triangle  is  equal  to  one- third  of  the  altitude  of 

I    \he  triangle. 

6.  The  quadrilateral  ABCD  is  inscribed  in  a  circle; 
AB,  DC  produced  meet  in  E:  prove  that  the  triangles 
ACE,  BDE,  and  also  the  triangles  ADE,  BCE,  are  mutu- 
ally equiangular. 

7.  The  quadrilateral  ABCD  is  inscribed  in  a  circle;  AB, 
DC  produced  meet  in  E,  and  BC,  AD  produced  meet  in 
F;  the  sides  AB,  BC,  CD  subtend  arcs  of  120°,  70°,  80°, 
respectively:  find  the  number  of  degrees  in  the  angles 
AED  and  AFB. 

8.  In  the  circumscribed  quadrilateral  ABCD,  the  angles 
A,  B,  C  are  110°,  95°,  80°,  respectively,  and  the  sides  AB, 
BC,  CD,  DA  touch  the  circumference  at  the  points  E,  F, 
G,  H  respectively:  find  the  number  of  degrees  in  each  an- 
gle of  the  quadrilateral  EFGH. 

/     9.  If  two  opposite  sides  of  an  inscribed  quadrilateral  are 
equal,  prove  that  the  other  two  sides  are  parallel. 

Problems  of  Construction". 

253.  Hitherto,  our  investigations  have  been  purely 
theoretical,  and  have  been  confined  to  the  demonstration  of 
certain  properties  of  figures,  assumed  to  exist,  satisfying 
certain  conditions  ;  and  our  figures  have  been  assumed  to 
be  constructed  under  these  conditions,  although  no  methods 
of  constructing  them  have  been  given.  Indeed,  the  precise 
construction  of  the  figures  was  not  necessary,  as  they  were 
required  only  as  aids  in  following  the  demonstration  of 
principles* 


BOOK  IL-PROBLEMS  OF  CONSTRUCTION.        105 

The  constructions  were  only  hypothetical.  We  now  ap- 
ply these  principles  to  determine  methods  by  means  of 
which  such  figures  are  to  be  approximately  drawn;  for, 
owing  to  the  imperfection  of  our  instruments,  the  ideal 
state  contemplated  in  theoretical  Geometry  cannot  be  at- 
[tained  by  them. 

The  determination  of  the  method  of  constructing  a  given 
^figure  with  given  instruments  is  called  di.  proUem  ;  and  the 
solution  oi  a  problem  requires  us  to  shoio  how  the  construc- 
tion can  be  affected  by  the  use  of  the  given  instruments, 
and  to  prove  that  the  construction  is  correct.  The  solution 
of  a  problem  depends  on  the  instruments  that  are  used. 
The  more  restricted  the  choice  of  instruments,  the  more 
limited  will  be  the  problems  which  can  be  solved  by  their 
use ;  and  the  more  difficult  will  be  the  solution  of  many 
that  are  thus  solved. 

It  IS  the  recognized  convention  of  Elementary  Geometry 
that  the  only  instruments  to  be  employed  are  the  ruler  and 
compasses,  with  the  use  of  which  the  student  should  be- 
come familiar.  The  ruler  is  used  for  drawing  and  produc- 
ing straight  lines,  and  the  compasses  for  describing  circles 
and  for  the  transference  of  distances;  the  straight  line  and 
the  circumference  being  the  only  lines  treated  of  in  this 
subject.  This  convention  is  embodied  in  the  three  postu- 
lates given  in  (45). 

Problems,  though  important  as  applications  of  geometric 
truths,  form  no  part  of  the  chain  of  connected  truths  em- 
bodied in  the  theorems  of  Geometry,  so  that,  though  they 
may  be  studied  advantageously  in*  connection  with  the 
theorems  on  which  they  directly  depend,  they  are  not  a 
necessary  part  of  the  pure  science  of  Geometry.* 

*  Elements  of  Plane  Geometry,  Association  for  the  improvement  of  Geometric 
Teaching,  p.  69. 


106 


PLANE  GEOMETRY, 


Proposition  2 1 .    Problem. 
254.  To  Used  a  given  finite  straight  line. 


I 


-^ 


-f=^-i- 


1 


\ 


I  / 


:d 


Given,  the  line  AB. 

Required,  to  bisect  it. 

Cons,  With  A  and  B  as  centres,  and  eqnal  radii  greater 
than  one-half  of  AB,  describe  two  arcs  cutting  each  other 
at  C  and  D. 

Join  CD  cutting  AB  at  E. 

Then  AB  is  bisected  at  E. 

Proof^mQQ  the  two  pts.  C  and  D  are  equallj  distant 

from  K  and  B,  the  st.  ImeUD  Is  jL  to  AB  at  its  mid- 

^HTe'pt:'       ~^  (67) 

Q.E.F. 
EXERCISES. 

^  1.  If  two  chords  intersect  at  right  angles  within  a  circle, 
prove  that  the  sum  of  tfae  opposite  intercepted  arcs  is  equal 
to  a  semi-circumference. 

2.  If  the  angles  A,  B,  C  of  a  circumscribed  quadrilateral 
ABCD  are  120°,  80°,  and  100°,  and  the  sides  AB,  BC,  CD, 
DA  touch  the  circumference  at  the  points  E,  F,  O,  H, 
find  the  number  of  degrees  in  each  angle  of  the  quadri- 
lateral E,  F,  G,  IT. 

3.  Divide  a  line  into  four  equal  parts, 


BOOK  Il.-PROBLEMS  OF  CONSTRUCTION.        107 


Proposition  22.    Problem. 
p    255.  To  bisect  a  given  arc  or  a  given  angle. 

(1)  Given,  the  arc  AB. 
Required,  to  bisect  it. 
Go7is.  Join  AB. 

With  A  and  B  as  centres,  and  with 
equal  radii,  describe  arcs  intersecting 
at  C  and  D. 

Draw  CD  cutting  the  arc  AB  at  E. 

Then  the  arc  AB  is  bisected  at  E. 

Proof,    Since  the  two  pts.  C  and  D  are  equally  distant 
_f  rom  A  and  B,  the  st.  line  CD  is  J_  to  the  chord  AB  at  its 
mid.  pt.  (67),  and  therefore  bisects  the  arc  (202). 

(2)  Give7i,  the  ZACB. 
Required,  to  bisect  it. 
Cons,   With  C  as  a  centre,  and  with  any 

radius,  describe  an  arc  cutting  CA  and  CB 
at  D  and  E. 

With  D  and  E  as  centres,  and  with  a 
radius  >  |  the  st.  line  DE,  describe  two 
arcs  intersecting  at  H. 

Join  CH. 

Then  CH  bisects  the  Z  ACB. 

Proof,  Join  DE.  Since  the  two  pts.^_and  E^  are 
equally  distant  fromC  and  H,  the  st.  line  CH  is  _L  to  the 
chord  DE  at  its  mid.  pt.  (67),  andlherefore  bisects  the  arc 
DE  and  the  Z'DCE  (202).  Q.e.f. 

EXERCISES. 

1.  Divide  an  arc  into  four  equal  parts. 

2.  Divide  an  angle  into  four  equal  parts. 


108  PLANE  GEOMETRY. 


Proposition  23.     Problem. 

256.  At  a  given  point  in  a  given  straight  line  to  draw 
a  perpendicular  to  that  line, 

Oiven,  the  pt.  C  in  the  line  AB. 

Required,  to  draw  from  C  a  i.  to  ">k 

AB.  j 

First  Method.  i 

D  I  E 

Cons,  With    0   as  a  centre,  and    ^     '  C  '     B 

with  any  radius,  describe  arcs  of  a  O 

cutting  AB  at  D  and  E. 


With  D  and  E  as  centres,  and  with  equal  radii  greater 
than  DC,  describe  two  arcs  cutting  each  other  at  H. 

Join  HC. 

Then  HC  is  the  required  _L. 

Proof.  Since  the  twojQts^  C  and  H  are  equally  distant 
from  D  an037the  stT  line  CH  is  _L  to  DE  at  its  mid, 
pt.  C.  (67). 

Second  Method. 

Cons.  With  any  pt.  0  without  AB  as  I  o 

a  centre,  and  with  the  distance  OC  as  ^^AE 

a  radius,  describe  a  Oce  cutting  AB  /'     |\ 

at  C  and  D.  ^.-'^ ^        |/ 

Join  DO,  and  produce  it  to  meet  A  D  \^             /'  C  B 

the  Oce  at  E.     Join  EC.  ,      "        ' 

Then  EC  is  the  required  _L. 

Proof,  Since  the  Z  ACE  is  inscribed  in  a  semicircle,  it 
is  a  rt.  Z  ;  .  •.  EC  is  _L  to  AB  at  C  (240).  q.e.f., 


BOOK  IL-PROBLEMS  OF  CONSTRUCTION.        l09 


Proposition  24.    Problem. 

257.  From  a  given  point  without  a  given  straight  line 
to  draw  a  perpendicular  to  that  line. 


Given,  the  pt.  0  and  the  line  AB. 

Required,  to  draw  a  _L  from  0  to  AB. 

Cons.  With  C  as  a  centre,  and  with  a  radius  sufficiently 
great,  describe  an  arc  cutting  AB  at  D  and  E. 

With  D  and  E  as  centres,  and  with  a  radius  >  ^  of  DE, 
describe  arcs  cutting  each  other  at  G. 

Join  CG,  cutting  AB  at  H. 

Then  OH  is  the  required  _L. 

Proof.  Since  the  two  pts.  C  and  G  are  equally  distant 
from  D  and  E,  the  st.  line  CG  is  J_  to  DEaUts  mid-pt.   (67) 

Q.E.F. 
EXERCISES. 

1.  In  an  indefinite  straight  line  AB  find  a  point  equally 
(     distant  from  two  given  points  which  are  not  lolh  on  AB. 

When  does  tliis  problem  not  admit  of  solution? 

2.  In  a  given  straight  line  MN"  find  a  point  P  such  that 
PA,  PB,  drawn  from  P  to  two  given  points  A,  B,  on  oppo- 
site sides  of  MN,  may  make  equal  angles  with  MN, 


no  PLANE  GEOMETRY. 


Proposition  25.    Problem. 

258.  At  a  given  point  on  a  given  straiglit  line  to  con- 
struct an  angle  equal  to  a  given  angle, 

E 


N 


/ 


A  MB 

Given,  the  pt.  A,  the  line  AB,  and  the  Z  0. 

Reqniredy  to  make  at  A  an  Z  =  Z  C. 

Co7is,  With  centre  C  and  any  radius,  describe  an  arc 
cutting  CD  and  CE  at  G  and  H.     Join  GH. 

With  centre  A  and  the  same  radius  CG,  describe  an  in- 
definite arc  MN. 

With  centre  M  and  radius  =  GH,  describe  an  arc  cutting 
the  arc  MN  at  F. 

Join  AF. 

Then  ZA=  ZC. 

Proof.  Since  chord  MN  =  chord  GH,  (Cons.) 

.  • .   arc  MN  =  arc  GH,  and  Z  A  =  Z  C. 
In  equal  Qs  equal  cJiords  subtend  egual  arcs,  and  equal  Zs  at  the 

centre  (198). 

Q.E.F, 

EXERCISES. 

1.  Given  an  angle,  to  construct  its  supplement, 

2.  Construct  an  angle  of  45°. 

3.  Construct  an  angle  of  22J°. 


BOOK  U.-FUOBLEMS  OF  CONSTRUCTION.        Ill 


Proposition  26.    Problem. 

259.  Two  angles  of  a  triangle  being  given  to  find  the 
third. 

Given,  Z  A  and  /  B  of  a  A . 

Required,  to  find  the  third  /  of 

theA.  ^   ^ 

H\ 
Cons,   Draw  the  indefinite  straight  \  ^^F 

line  CD.     At  any  pt.  E  in  this  line,     ^^v^-rJll 

makeZDEFn:  ZB,  C  E  D 

and  zCEHrr  zA.  .    (258) 

Then  Z  FEH  is  the  Z  required. 

Proof.   Since  Z  CEH  +  Z  HEF  +  Z  FED  =  2  rt.  Z  s,  (53) 

and  ZA  +  zB  +  requiredZ  =  2  rt.   Zs,        (97) 

and  Z  A  =:  z  CEH,  and  Z  B  =  Z  FED,      (Cons.) 

.  • .  Z  HEF  is  the  Z  required.  q.e.f. 


Proposition  27.    Problem. 

260.  Through  a  given  point  to  draio  a  straight  line 
parallel  to  a  given  straight  line. 

Given,  the  pt.  A  and  the  line  BC.  /^ 

Required,  to   draw  through   A  a                      /' 
line  II  to  BC.  / 

Cons.  In  BC  take  any  pt.   D,  and       B         D  C 

join  DA. 

At  pt.  A  make  Z  DAE  =  Z  ADC.  (258) 

Then  '"   EAF  is  ||  to  BC. 

Proof      Since  Z  DAE  =  Z  ADC,  (Cons. ) 

.-.  EFislltoBC, 

being  alt.-iiit.  As.  (75).  q.E.F, 


112  PLANE  GEOMETRY. 


Proposition  28.    Problem. 

261.  Oiven  tivo  sides  and  the  i?icluded  angle  of  a  tri- 
angle, to  construct  the  triangle. 

Given,  two  sides  a,  h,  and  the  in- 
cluded Z  A. 

Required,  to  construct  the  A . 

Cons,     Draw  the  line  AB  =  a. 

At  A  make  the  /  BAD=  /  A.  (258) 

On  AD  take  AC  =  b. 

Join  CB. 

Then  ABC  is  the  A  required. 

Proof,    To  be  supplied  by  the  student.  q.e.f. 

Proposition  29.    Problem. 

262.  Given  a  side  and  the  tivo  adjacent  angles  of  a 
triangle,  to  construct  the  triangle. 

Given,   the   side   a  and  the   adj. 
ZsA,  B. 

Required,  to  construct  the  A .  I    a 

Cons.  Draw  the  line  AB  =  a.  C/'^^^^ 

At  A  make  the  Z  BAD  =  Z  A.  (258)      '/ ^Jl^^--:. 

At  B  make  the  Z  ABE=  Z  B.  (258)      ^  a  B 

The  lines  AD  and  BE  will  intersect  at  some  pt.  C. 
Then  ABC  is  the  A  required. 
Proof.     To  be  supplied  by  the  student.  q.e.f. 

263.  ScH.  1.  Since  the  third  angle  of  a  triangle  can  be 
found  when  two  angles  are  given  (259),  therefore,  if  a  side 
and  any  two  angles  of  a  triangle  are  given,  the  triangle 
may  always  be  constructed  (262). 

264.  ScH.  2.  This  problem  is  possible  only  when  the 
two  given  angles  are  together  less  than  two  right  angles. 


BOOK  IL-PliOBLEMS  OF  CONSTRUCTION.        U3 


Proposition  30.    Problem. 

265.  Given  the  three  sides  of  a  triangle  to  construct  the 
triangle. 


/     \ 
c/  \b 

/  \ 


/- \ 

B  a  C 


Given,  the  three  sides  a,  h,  c. 

Required,  to  construct  the  A . 

Cons.  Draw  the  line  BC  =  «. 

With  C  as  a  centre,  and  a  radius  =  h,  describe  an  arc. 

With  B  as  a  centre,  and  a  radius  =  c,  describe  an  arc, 
cutting  the  former  arc  at  A. 

Join  BA  and  CA. 

Then  ABC  is  the  A  required. 

Proof.  To  be  supplied  by  the  student. 

Q.E.F. 
A  EXERCISES. 

1.  Show  how  the  construction  fails  if  one  side  is  greater 
than  the  sum  of  the  other  two. 

2.  Construct  an  equilateral  triangle  on  a  given  base. 


/, 

0  — 

A. 

^ 

^ 

A' 

'^, 

a--"E 

/^'^x 

b^ 

-> 

/      1      \ 

1 

\  a 

^-"/ 

la 

1 

NT 

-V — 

.^=t^_ 

\ 

B^N. 

H 

,• 

114  PLANE  GEOMETRY, 


Proposition  31 .    Problem. 

266.  To  construct  a  triangle  having  given  two  sides  and 

the  angle  opposite  one  of  them. 

Given,  the  sides  a,  h,  and  /  A 
opp.  a. 

Required,  to  construct  the  A . 

Case  I.  When  a  <h,  and 
Z.  A  is  acute. 

Cons,  At  pt.  A  on  the  in- 
definite st.  line  AD,  make  the 

ZDAE=ZA.  (258) ,__ 

A     rx            H  /B'D 

On  AE  take  AG  =  h.  ^-^.... ,../'' 

AVith  C  as  a  centre,  and  radius 
=  a,  describe  an  arc  cutting  AD  at  B  and  B'. 

Join  CB  and  CB'. 

Then,  either  ABC  or  AB'C  is  the  A  required,  since  each 
satisfies  the  given  conditions;  and  the  problem  has  two 
sohitions. 

This  is  called  the  ambiguous  case. 

When  the  side  a  —  the  J_  CH,  there  is  but  one  solution : 
the  rt.  A  ACH,  since  the  arc  described  with  centre  C  and 
radius  a  touches  AD. 

When  the  side  a  <  CH  the  problem  is  impossible,  since 
the  arc  described  with  centre  C  and  radius  a  does  not  cut 
or  touch  AD. 

Hence,  the  arc  described  with  centre  C  and  radius  a  will 
meet  AD  in  two  pts.  B  and  B'  equidistant  from  H,  or  in 
one  pt.  H,  or  in  no  pt.  whatever,  according  as  ^  >,  =,  or 
<  CH.  Therefore,  there  may  be  two  As,  one  A;  or  none 
•it  all. 


BOOK  II.— PROBLEMS  OF  CONSTRUCTION.        115 

Case  IL    When  a  ^h,  and  /_K  is  acute. 

In   this  case  the   arc    described  /^ 

with    centre  C  and  radius  a,  cuts  ^^^ 

AD  at  the  two  pts.  A  and  B;  and  b/        \a 

hence  there  is  but  one  solution:  the       ^  /  \  / 

isosceles  A  ABU.  ^i^Z,  ~~yB"D 

Case  III.   When  a>h, 

"When  Z  A  is  acute  the  arc  de-  /^ 

scribed  with  centre  C  and  radius  a  ^'^^ 

cuts  AD  in  B  and  B',  the  latter  pt.  ^^/b    ^^"^X 

at  the  left  of  A  in  DA  produced.    gV — -^ "/q"^ 

Then   ABC    is    the    A    required,  X._  ,,-•'' 

since  it  satisfies  the  given  condi- 
tions, and  there  is  only  one  solution,  for  the  A  AB'C  does 
not  contain  the  given  /  A. 

When  Z  A  is  ohtu.se,  as  /  CAB',  there  is  only  one  solu- 
tion: the  A  AB'C,  since  the  A  ABC  does  not  contain  the 
given  obtuse  Z . 

When  Z  A  is  right  there  are  two  C 

equal  rt.  As:  ABC  and  AB'C.  /|\ 

When  Z  A  is  right  or  obtuse,  and  V  ^i    V 

r;  =:  or  <  h,  the   problem  is  impos-    __:x^ J_ ^ 

sible;   for  the  side  opposite  the  right        B' •-...... A --"b 

or  the  obtuse  Z  is  the  greatest  side  of  the  A  (117). 

Q.E.F. 
EXERCISES. 

-  1.  Construct  a  right  triangle  having  given  the  hypote- 
nuse and  one  side. 

2.  Construct  au  isosceles  right  triangle  on  a  given  straight 
line  as  hypotenuse. 


-^--D 


116  FhANE  GEOMETRY, 


Proposition  32.    Problem. 

267.  Given  two  adjacent  sides  and  the  included  angle 
of  a  parallelograniy  to  construct  the  parallelogram^. 

Given,  the  sides  a,  h,  and  the 
ZA.  

Required,  to  construct  the  C7.  ^ 

Cons.  Draw  AB  =  «. 

At  A  make  the  Z  B AE  =  Z  A. 


On  AE  take  AC  =  h  /E^ 

With  C  as  a  centre,  and  a  ra-  c/^^  ^^ 

dius  =  a,  describe  an  arc.                      b/  / 

With  B  as  a  centre,  and  a  ra-       / / 

dins  =  b,  describe  an  arc  cutting      A  ~a  b 

the  first  at  D. 

Join  CD,  BD. 

Then  ABCD  is  the  required  OJ, 

Proof.  Since  AB  =  CD  and  AC  =  BD  (Cons.),  the 
figure  is  a  /Z7  (132),  and  it  is  the  one  required;  for  two  /Z7s 
are  equal  when  they  have  two  adj.  sides  and  the  included 
Z  equal  each  to  each  (135). 

Q.E.F. 
EXERCISES. 

>-  1.  Construct  a  square  upon  a  given  straight  line. 

2.  Construct  a  parallelogram,  having  given  two  adjacent 
sides  and  one  diagonal. 
7\3.  Construct  a  rhombus,  having  given  the  two  diagonals. 

4.  On  a  given  straight  line  as  hypotenuse,  construct  a 
[N^ight  triangle  having  one  of  its  acute  angles  double  the 
other. 


BOOK  IL— PROBLEMS  OF  CONSTUUCTION,        117 


Proposition  33.    Problem. 


268.  To  draw  a  tangent  to  a  given  circle  from  a  given 
point  either  on  or  witlwut  the  circumfere7ice» 

Given,  the  O  BCD,  and  the  pt.  A. 
Jieqtiired,  to  draw  from  A  a  tangent   ^i 
to  the  O  BCD.  | 

Case  I.  When  the  pt.  A  is  ifi  the  Qce. 
Cons.  Draw  the  radius  OA. 
At  A  draw  AE  _L  to  AO.  (256) 

Then  AE  is  the  tangent  required.   (209) 

Case  II.    When  the  pt.  A  is  without  the  O, 
Cons.  Join  OA;  bisect  it  at  E. 

Witli  centre  E  and  radius  EO,  describe 
a  Oce,  cutting  the  O  BCD  at  B,  D. 

Join  AB,  AD. 

Then  either  AB  or  AD  is  the  tangent 
required. 

Proof.  Join  OB,  OD. 

Z ABO  =  rt.  Z ,  and  Z  ADO  =  rt.  Z, 

being  inscribed  in  a  semicircle  (240). 

.*.  AB  and  AD  are  each  tangent  to  the  O  at  B,  D, 
being  i  to  the  radius  at  its  extremity  (209). 

Q.E.F. 

ScH.  When  the  pt.  A  is  without  the  O,  two  tangents 
may  always  be  drawn,  and  they  wdl  be  equal;  for 
rt.A  AOB  =  rt.A  AOD.  (110) 


118 


PLANE  Geometry, 


Proposition  34.    Problem. 

269.  To  inscribe  a  circle  in  a  given  triangle. 
Given,  the  A  ABC. 
Required,  to   inscribe  u  0  in 

A  ABC. 

Cons.  Bisect  the  Z  s  A,  B  by  the 
lines  AO,  BO  meeting  at  0. 

From  0  draw  OD  i.  to  AB. 

With  centre  0  and  radius  OD, 
describe  the  O  DEF. 

The  O  DEF  is  the  required  o. 

Proof,  Since  the  pt.  of  intersection  of  the  bisectors  of 
the  angles  of  a  A  is  equally  distant  from  the  three  sides  of 
the  A,  (1G7) 

.-.the  J_s  OD,  OE,  OF  are  equal. 

.•.  a  O  described  with  centre  0  and  radius  OD  will  be 
tangent  to  the  three  sides  of  the  A  at  the  pts.  D,  E,  F,  and 
be  inscribed  in  it.  (191) 

Q.E.F. 

270.  ScH.  The  centre  of  the  circle  inscribed  in  a  tri- 
angle is  sometimes  called  its  in-centre. 

Note.— The  bisectors  of  the  angles  of  a  triangle  are  concurrent,  the  point  of 
intersection  being  the  centre  of  the  circle  inscribed  in  the  triangle. 

271.  Dep.  If  the  sides  of  a  triangle  are  produced  and 
the  exterior  angles  are  bisected,  the  intersections  of  the 
bisectors  are  the  centres  of  three  circles,  each  of  which  is 
tangent  to  one  side  of  the  triangle  and  the  other  two  sides 
produced.     These  three  circles  are  called  escribed  circles. 


BOOK  II.— PROBLEMS  OF  CONSTRUCTION.        119 


Proposition  35.    Problem. 

^272.  To  draw  an  escribed  circle  of  a  given  triangle. 


/ 


Given,  the  A  ABC. 

Required,  to  describe  a  O 
touching  AB,  and  C  A,  CB  pro- 
duced. 

Co7is.  Bisect  the  Z  s  BAE, 
ABD  by  the  lines  AO,  BO, 
which  intersect  at  0.        (255) 

From  0  draw  OG,  OH,  OK 
1  to  CE,  AB,  CD. 

With   centre  0  and  radius    ^    \ 
Oa,  describe  the  O  OHK. 

The  O  GHK  is  the  required 
O. 

Proof,  Since  the  pt.  0  is  on  the  bisector  of  Z  BAE,  it  is 
equally  distant  from  BA  and  AE.  (160) 

.•.0G  =  0H. 

Similarly,  OH  =  OK. 

.  • .  the  _Ls  00,  OH,  OK  are  equal. 

.  • .  a  O  described  with  centre  0  and  radius  00  will  touch 
CE,  AB,  CD  at  the  pts.  G,  H,  K. 

.  • .  the  O  GHK  is  an  escribed  O  of  the  A  ABC.        q.e.f. 

ScH,  In  the  same  manner  the  centres  0',  0"  of  the  other 
two  escribed  Os  may  be  found. 

Therefore  there  are  in  general  four  circles  tangent  to  three 
intersecting  straight  lines. 


120  PLANE  GEOMETRY. 


Proposition  36.    Problem. 

273.  To  circumscribe  a  circle  about  a  given  triangle. 

Given,  the  A  ABC.  . ^ 

Required,  to  circumscribe  aO  about  the       /'^         /\  \ 

A  ABC.  /       d/    \\ 

Cons.  Draw  DO,  EO  _L  to  AC,  AB  at  \  />\  \j 
tlieir  mid.  pts.  intersecting  at  0  (284).  a\ e /  ^ 

With  0  as  a  centre,  and  a  radius  =  OA,  ^^ -^^ 

describe  a  O. 

This  O  is  the  one  required  passing  through  the  vertices 
A,  B,  C. 

Proof.  Since  the  pt.  0  is  in  the  _L  bisectors  OD,  OE  of 
AC,  AB,  it  is  equally  distant  from  the  pts.  A,  B,  C. 

JSver^  pt.  ill  Hie  ±  bisector  of  a  line  is  equidistant  from  tfie  extremities 
of  the  line  (66). 

.  • .  a  O  described  with  centre  0  and  radius  OA  must  pass 
through  the  pts.  A,  B,  C,  and  is  .*.  circumscribed  about 
the  A  ABC.  Q.E.F. 

274.  ScH.  This  construction  is  the  same  as  that  of  de- 
scribing a  circumference  through  any  three  given  points 
not  in  the  same  straight  line,  or  of  finding  the  centre  of  a 
given  circle,  or  of  a  given  arc. 

Note.— Since  the  perpendicular  bisector  of  a  chord  passes  through  the  centre 
of  the  circle  <202),  therefore, 

The  perpendicular  bisectors  of  the  sides  of  a  triangle  are  concurrent,  the 
point  of  intersection  being  the  centre  of  the  circle  circumscribed  about  the 
triangle. 

The  centre  of  the  circle  circumscribed  about  a  triangle  is 
sometimes  called  its  circum-centre. 

EXERCISE. 

Prove  (1)  if  the  given  triangle  be  acute-angled,  the  centi(^ 
of  the  circumscribe!  circle  falls  within  it;  (2)  if  it  be  a 
right  triangle,  the  centre  falls  on  the  hypotenuse;  (3)  if  it 
be  an  obtuse-angled  triangle,  the  centre  falls  without  the 
triangle. 


BOOK  II.-PROBLEMS  OF  CONSTRUCTION,        121 


Proposition  37.    Problem. 

275.   On  a  given  straight  line,  to  describe  a  segment 
which  shall  contain  a  given  angle* 

Given,  the  st.  line  AB. 

Required,  to  describe  on  AB  a  seg-       ^'^      '"^^^ 
ment  containing  /  C.  ^Z  x\^ 

Co7is,  Make  /  BAD  =  Z  C.    (2^8)    /  q/ ' 


From  A  draw  AH  J.  to  AD.     (256)    \ 


\y    !   \V 


Bisect  AB  m  E.  (254)     )^ ^ f^ 

From  E  draw  EO  _L  to  AB  meeting         \' 

AH  at  0.                                       (256)  '"^^ 

With  0  as  a  centre,  and  OA  as  a  C^ 

radius,  describe  the  Oce  AHBF. 

The  segment  AHB  is  the  segment  required. 

Proof,  Join  OB  and  BH.     Since  0  is  in  the  J_  bisector  of 
AB,  it  is  equally  distant  from  the  pts.  A,  B.  (66) 

.•.  a  O  described  with  centre  0  and  radius  0  A  must  pass 
through  B. 

Also,  since  AD  is  _L  to  AH  at  its  extremity, 

.-.  AD  is  tangent  to  the  O  at  A.  (209) 

Since  AB  is  a  chord  through  A, 

.\   Z  BAD  is  measured  by  i  arc  AFB.  (243) 

But  f  AHB  is  measured  by  J  arc  AFB.  (238) 

.'.  any  Z  in  the  segment  AHB  =  zC.         (Ax.  1) 

Q.E.F. 

Note.— In  the  particular  case  when  the  given  angle  C  is  a  right  angle,  the  seg- 
ment required  will  be  the  semicircle  described  on  the  given  st.  line  AB. 


129 


PLANE  GEOMETHT. 


Proposition  38.     Problem. 


1 


S76.   To  draw  a  common  tangent  to  two  given  circles. 

Given,  the  Os  AR,  BS,  and 
let  AR  >  BS. 

(1)  Required,  to  draw  an  ex- 
terior common  tangent  to  the 
two  Os. 

Cons,  With  centre  A  and  ra- 
dius =  the  difference  of  the 
radii  of  the  two  Os,  describe 
a  O. 

From  B  draw  BC  tangent  to  this  O-  (268) 

Join  AC,  and  produce  it  to  meet  the  Oce  of  the  given 
O  in  D. 

Draw  BE  II  to  AD,  and  join  DE. 

DE  is  a  common  tangent  to  the  two  given  Os. 


Proof. 


But 


But 


Since  AC  =  AD  -  BE, 
.-.  CD  =  BE. 

CD  is  II  to  BE, 
.-.  BCDE  is  a  CJ, 
.-.  DE  is  =:  and  II  to  CB. 


(Cons.) 

(Cons.) 
(133) 
(124) 


ZACB=:art.  Z. 
A  tang,  to  a  q  at  any  pt.  is  L  to  the  radius  at  thai  pi.  (210). 

.*.  BCDE  is  a  rectangle.  (77) 

.-.    ZD  =  ZE  =  art.  Z. 
.-.  DE  is  a  tangent  to  both  Os.  q.e.f. 

Since  two  tangents  can  be  drawn  from  B  to  the  O  AR, 
therefore  two  common  tangents  may  always  be  drawn  to  the 
given  Os.     These  are  called  the  direct  common  tangents. 

AVhen  the  given  Os  are  external  to  each  other  and  do 
not  intersect,  two  more  common  tangents  may  be  drawn, 
called  the  transverse  co7mnon  tangents. 


BOOK  1L—HXERCI8ES.     TltEOliEMS. 


123 


(2)  Required,  to  draw  the 
transverse  pair  of  common 
tangents. 

Cons,  and  proof.  With  centre  A  and 
radius  =  the  sum  of  the  radii  of  the  two 
OS,  describe  a  ©,  and  complete  the  con- 
struction, and  proof,  as  in  (1). 


exercises. 
Theobems. 


1.  If  a  straight  line  cut  two  concentric  circles,  the  parts 
of  it  intercepted  between  the  two  circumferences  are  equal. 

2.  If  one  circle  touch  another  internally  at  P,  prove  that 
the  straight  line  joining  the  extremities  of  two  parallel 
diameters  of  the  circles,  towards  the  same  parts,  passes 
through  P. 

3.  In  Ex.  2,  if  a  chord  AB  of  the  larger  circle  touches 
the  smaller  one  at  C,  prove  that  PC  bisects  the  angle  APB. 

4.  If  two  circles  touch  externally  at  P,  prove  that  the 
straight  line  joining  the  extremities  of  two  parallel  diame- 
ters towards  opposite  parts,  passes  through  P. 

5.  Two  circles  with  centres  A  and  B  touch  each  other 
externally,  and  both  of  them  touch  another  circle  with 
centre  0  internally  :  show  that  the  perimeter  of  the  tri- 
angle AOB  is  equal  to  the  diameter  of  the  third  circle. 

6.  In  two  concentric  circles  any  chord  of  the  outer  circle 
which  touches  the  inner,  is  bisected  at  the  point  of  conkict. 

7.  If  three  circles  touch  one  another  externally  in  P,  Q,  R, 
and  the  chords  PQ,  PK  of  two  of  the  circles  be  produced 
to  meet  the  third  circle  again  in  S,  T,  prove  that  ST  is  a 
diameter. 

8.  Points  P,  Q,  R  on  a  circle,  whose  centre  is  0,  are 
joined  ;  OM,  ON  are  drawn  perpendicular  to  PQ,  PR  re- 
spectively :    join  MN,  and  show  that  if  the  angle  OMN 


124  PLANE  OEOMETRT. 

is  greater  than  ONM,  then  the  angle  PRQ  is  greater  than 
PQR. 

9.  A  circle  is  described  on  the  radius  of  another  circle 
as  diameter,  and  two  chords  of  the  larger  circle  are  drawn, 
one  through  the  centre  of  the  less  at  right  angles  to  the 
common  diameter,  and  the  other  at  right  angles  to  the  first 
through  the  point  where  it  cuts  the  less  circle.  Show  that 
these  two  chords  have  their  greater  segments  equal  to  each 
other  and  their  less  segments  equal  to  each  other. 

10.  0  is  the  centre  of  a  circle,  P  is  any  point  in  its  cir- 
cumference, PN  a  perpendicular  on  a  fixed  diameter :  show 
that  the  straight  line  which  bisects  the  angle  OPN  always 
passes  through  one  or  the  other  of  two  fixed  points  on  the 
circumference. 

11.  Two  tangents  are  drawn  to  a  circle  at  the  opposite 
extremities  of  a  diameter,  and  intercept  from  a  third  tan- 
gent a  portion  AB:  if  C  be  the  centre  of  the  circle  show 
that  ACB  is  a  right  angle. 

12.  A  straight  line  touches  a  circle  at  A,  and  from  any 
point  P,  in  the  tangent,  PB  is  drawn  meeting  the  circle 
at  B  so  that  PB  is  equal  to  PA:  prove  that  PB  touches  the 
circle. 

13.  OC  is  dmwn  from  the  centre  0  of  a  circle  perpendic- 
ular to  a  chord  AB :  prove  that  the  tangents  at  A,  B  inter- 
sect in  OC  produced. 

14.  TA,  TB  are  tangents  to  a  circle,  whose  centre  is  0 ; 
from  a  point  P  on  the  circumference  a  tangent  is  drawn 
cutting  TA,  TB  or  those  produced  in  0,  D:  prove  that  the 
angje  COD  is  half  the  angle  AOB. 

15.  AB  is  the  diameter  and  C  the  centre  of  a  semicircle: 
show  that  0  the  centre  of  any  circle  inscribed  in  the  semi- 
circle is  equidistant  from  C  and  from  the  tangent  to  the 
semicircle  parallel  to  AB. 

16.  If  from  any  point  without  a  circle  straight  lines  be 
drawn  touching  it,  the  angle  contained  by  the  tangents  is 
double  the  angle  contained  by  the  straight  line  joining  the 


I 


BOOK  IL-EXEIWISES.     TIIEOllEMS.  125 


points  of  contact  and  the  diameter  drawn  through  one  of 
them. 

17.  C  is  the  centre  of  a  given  circle,  CA  a  radius,  B  a 
point  on  a  radius  at  right  angles  to  CA;  join  AB  and  pro- 
duce it  to  meet  the  circle  again  at  D,  and  let  the  tangent 
at  D  meet  CB  produced  at  E:  show  that  BDE  is  an  isosceles 
triangle. 

18.  Let  the  diameter  BA  of  a  circle  be  produced  to  P,  so 
that  AP  equals  the  radius;  through  A  draw  the  tangent 
AED,  and  from  P  di-aw  PEC  touching  the  circle  at  0  and 
meeting  the  former  tangent  at  E;  join  BO  and  produce  it 
to  meet  AED  at  D:  then  will  the  triangle  DEC  be  equi- 
lateral. 

Let  O  be  the  centre  of  the  given  0.    Produce  OC  to  a  pt.  F  so  that  CF  =  CO: 
compare  as  PCO,  PCF,  etc. 

19.  APB  is  a  fixed  chord  passing  through  P,  a  point  of 
intersection  of  two  circles  AQP,  PBR;  and  QPR  is  any 
other  chord  of  the  circles  passing  through  P:  show  that 
AQ  and  RB  when  produced  meet  at  a  constant  angle. 

20.  Two  circles  whose  centres  are  A  and  B  touch  exter- 
nally at  C;  the  common  tangent  at  C  meets  another  com- 
mon tangent  DE  at  F:  prove  that  (1)  CF,  DF,  FE  are 
equal;  (2)  each  of  the  angles  AFB,  DCE  is  a  right  angle; 
(3)  DE  touches  the  circle  described  on  AB  as  diameter. 

21.  The  diagonals  AC,  BD  of  a  quadrilateral  ABCD  in- 
scribed in  a  circle  intersect  at  right  angles  at  P:  prove  that 
the  straight  line  drawn  from  P  to  the  middle  point  of  one 
of  the  sides  of  the  quadrilateral  is  perpendicular  to  the  op- 
posite sides. 

Bisect  AB  in  E,  produce  EP  to  meet  CD  in  F,  etc. 

22.  If  a  side  of  a  quadrilateral  inscribed  in  a  circle  be 
produced,  the  Exterior  angle  is  equal  to  the  interior  and 
opposite  angle:  and  conversely,  if  the  exterior  angle  of  a 
quadrilateral  made  by  any  side  and  the  adjacent  side  pro- 
duced be  equal  to  the  interior  and  opposite  angle,  a  circle 
can  be  described  about  the  quadrilateral, 


126  PLANE  GEOMETRY. 


^ 


23.  ABOD  is  a  quadrilateral  inscribed  in  a  circle  with 
centre  0.  If  the  angles  BAD,  BOD  are  together  equal  to 
two  right  angles,  prove  that  the  angle  BCD  is  two-thirds 
of  two  right  angles. 

24.  If  two  pairs  of  opposite  sides  of  a  hexagon  inscribed 
in  a  circle  are  parallel,  the  third  pair  of  opposite  sides  are 
parallel. 

See  (342). 

25.  A  circle  is  described  passing  through  the  ends  of  the 
base  of  a  given  triangle:  prove  that  the  straiglit  line  join- 
ing the  points,  in  which  it  cuts  the  sides  or  sides  produced, 
is  parallel  to  a  fixed  straight  line. 

26.  In  the  semicircle  ABODE,  the  chord  BD  which  is 
parallel  to  the  diameter  AE  bisects  the  radius  00  at  right 
angles :  prove  that  the  arc  BO  is  double  the  arc  AB. 

Join  BO,  CD;  let  OC  cut  BD  in  F;  BF  =  FD,  etc. 

27.  The  straight  lines  which  bisect  the  vertical  angles  of 
all  triangles  on  the  same  base,  on  the  same  side  of  it,  and 
with  the  same  vertical  angle,  are  concurrent. 

28.  AB,  AO  are  chords  of  a  circle:  show  that  the  straight 
line,  which  joins  the  middle  points  of  the  arcs  AB,  AO, 
cuts  off  equal  portions  of  the  chords. 

Let  D,  E  be  the  mid  pts.  of  arcs  AB,  AC;  take  D,  E  on  opp.  sides  of  diam. 
through  A,  etc. 

29.  BAO,  BA'O  are  two  angles  in  the  same  segment  of 
a  circle;  AP,  A'P'  are  drawn  making  the  angles  BAP, 
BA'P'  equal  to  the  angles  BOA,  BOA'  respectively :  prove 
that  AP  is  parallel  to  A'P'. 

30.  OD  is  a  chord  of  a  circle  at  right  angles  to  the  diam- 
eter x\B ;  E  is  any  j^oint  in  the  arc  BO ;  AE  cuts  OD  in  F : 
prove  that  the  angles  DFE,  AOE  are  equal. 

31.  Prove  that  two  of  the  straight  lines  which  join  tlie 
ends  of  two  equal  chords  are  parallel,  and  that  the  other 
two  are  equal. 

Let  the  equal  chds.  AB,  CD  be  joined  towards  the  same  parts  by  BC,  AD; 
towards  opp.  parts  by  AC,  BD;  ,•.  chd.  AB  =  chd.  CD,  etc. 


BOOK  IL— EXERCISES.     LOCI.  127 

32.  ABCD  is  a  quadrilateral  inscribed  in  a  circle.  Find 
the  relation  between  the  sides  in  order  that  AC  may  \)isect 
the  angle  BAD,  and  BD  bisect  the  angle  ABC. 

/DAC  =  /CAB, .-.  arc  DC  =  arc  CB,  .'.  CB  =CD,  etc. 

33.  Two  circles  touch  internally  at  0,  and  a  line  is  dra^n 
cutting  one  of  the  circles  in  P,  P',  the  other  in  Q,  Q' :  show 

»iat  PQ,  P'Q'  subtend  equal  angles  at  0. 
Draw  tang.  OB,  B  being  towards  P,  Q,  etc. 
34.  Two  circles  touch  externally  at  A;  the  tangent  at  B 
to  one  of  them  cuts  the  other  in  C,  D :  prove  that  BC  and 
BD  subtend  supplementary  angles  at  A. 

Draw  common  tang.  AE  meeting  BC  in  E,  Z  EAB  =  Z  EBA,  etc. 

35.  If  the  opposite  pairs  of  sides  of  a  quadrilateral  in  a 
circle  be  produced  to  meet,  and  the  angles  so  formed  be 
bisected,  the  bisectors  are  at  right  angles  to  each  other. 

Let  ABCD  be  a  cyclic  quadl. ;  let  AD,  BC  meet  in  E,  and  AB,  DC  in  F,  etc. 

36.  If  from  any  point  on  the  circumference  of  the  circle 
circumscribing  a  triangle,  perpendiculars  be  drawn  to  the 
sides,  show  that  the  feet  of  these  perpendiculars  are  col- 
liiiear  (1()4). 

Let  P  be  any  pt.  in  the  G  ce  of  the  O  circumscribing  the  a  ABC.    From  P  draw 
PL  ±  to  BC,  PN  ±  to  AB,  etc. 

37.  In  any  triangle  ABC,  if  0  be  the  orthocentre  (171), 
and  L,  M,  N  the  feet  of  the  perpendiculars,  the  circle  de- 
scribed through  L,  M,  N  will  (1)  bisect  OA,  OB,  OC,  and 
(2)  will  also  pass  through  the  middle  points  D,  E,  F  of  the 
sides  of  the  triangle. 

Loci. 

38.  Find  the  locus  of  a  point  at  a  given  radial  distance 
from  the  circumference  of  a  given  circle. 

See  (157). 

39.  On  a  given  base  as  hypotenuse  right  triangles  are  de- 
scribed: find  the  locus  of  their  vertices. 

See  (240). 


^ 


128  PLANE  GEOMETRY. 

40.  Find  the  locus  of  the  centre  of  a  circle  which  passes 
through  two  given  points. 

See  (159)  and  (202). 

41.  Find  the  locus  of  the  centi-e  of  a  circle  which  touches 
a  given  straight  line  at  a  given  point,  or  which  touches  a 
given  circle  at  a  given  point. 

42.  Find  the  locus  of  the  centre  of  a  circle  which  touches 
a  given  straight  line,  and  has  a  given  radius. 

43.  Find  the  locus  of  the  centre  of  a  circle  which  touches 
a  given  circle,  and  has  a  given  radius. 

44.  Find  the  locus  of  the  centre  of  a  circle  which  touches 
two  given  straight  lines. 

See  (162). 

45.  Prove  that  the  locus  of  the  middle  points  of  cqu^il 
chords  in  a  given  circle  is  another  concentric  circle. 

4G.  Prove  that  the  locus  of  the  middle  points  of  all 
chords  drawn  through  a  given  point  in  a  given  circum- 
ference, is  a  circle  passing  through  the  given  point  and 
having  the  radius  of  the  given  circle  for  a  diameter. 

47.  Prove  that  the  locus  of  the  vertex  of  a  triangle  having 
a  given  base  and  a  given  vertical  angle,  is  a  circle. 

48.  A  straight  line  moves  so  that  its  ends  constantly  touch 
two  fixed  lines  at  right  angles  to  each  other  :  prove  that 
the  locus  of  its  middle  point  is  a  circle:  find  its  centre  and 
radius. 

49.  A  and  B  are  fixed  points,  and  C  is  any  point  on  a 
circle ;  AC  is  produced  to  D  so  that  CD  is  always  equal  to 
CB :  find  the  locus  of  D. 

CD  =  CB,  .-.  ZCDB  =  ZCBD,  etc. 

50.  Through  one  of  the  points  of  intersection  of  two 
fixed  circles,  centres  A  and  B,  a  chord  is'  drawn  meeting 
the  first  circle  in  P  and  the  other  in  Q.  Find  the  locus  of 
the  point  of  intersection  of  PA  and  QB. 

Let  D  be  the  intersection  of  0s.    (1)  Let  P,  Q  be  each  on  the  arc  outside  the 
other  O ;  let  PA,  QB  meet  in  R,  etc. 


.    BOOK  IL— EXERCISES.     PROBLEMS.  129 

IH  51.  ABC  is  a  triangle  inscribed  in  a  circle,  P  any  point 
in  the  circumference  of  the  circle,  and  Q  is  a  point  in  PC 
such  that  the  angle  QBC  is  equal  to  the  angle  PBA.  Prove 
that  the  locus  of  Q  is  a  circle. 

If  P  is  in  arc  AB,  because  ZQBC  =  ZPBA,  and  ZQCB  =  /PAB,  .-.  ZBQC  = 
^^^  BPA,  etc. 

^^m  52.  Find  the  locus  of  the  middle  points  of  all  chords  of 
^^Rcircle  which  pass  through  a  given  point. 

^^KiBt  P  be  the  given  iJt.,  O  the  cent,  of  the  O,  AB  a  chd.  through  P,  etc. 

^^  53.  From  any  point  P  on  the  circumference  of  a  circle 
circumscribing  a  triangle  ABC,  perpendiculars  PD,  PE  are 
let  fall  on  the  sides  AB,  AC  :  prove  that  the  locus  of  the 
centre  of  the  circle  circumscribing  the  triangle  PDE  is  a 
circle. 


Problems. 

If  the  construction  of  a  problem  does  not  readily  appear 
to  the  student,  he  should  present  a  careful  analysis,  that  is, 
a  course  of  reasoning  by  which  the  solution  may  be  arrived 
at.  In  an  analysis  we  generally  begin  by  supposing  the 
problems  solved,  and  constructing  the  figure  accordingly. 
We  then  reason  out  the  conditions  to  be  fulfilled,  study  the 
relations  of  the  lines,  angles,  etc.,  in  the  figure,  draw  aux- 
iliary lines  parallel  or  perpendicular,  as  the  case  may  require, 
join  given  points  or  points  assumed  in  the  solution,  describe 
circles  when  necessary,  and  endeavor  to  discover  the  de- 
pendence of  the  problem  upon  previously  solved  problems 
or  theorems.  The  reverse  process,  or  synthesis,  then  fur- 
nishes a  construction  of  the  problem.  The  analysis  will 
furnish  exercise  for  the  student^s  ingenuity  in  drawing 
useful  auxiliary  lines.  The  synthesis  will  furnish  exercise 
for  his  invention  in  combining  the  different  steps  suggested 
by  the  analysis  in  the  simplest  form. 


130  PLANE  GEOMETRY. 

54.  In  a  given  circle  to  inscribe  a  triangle  equiangular  to 
a  given  triangle. 

Given,  the  O  ABO,  and  the 
aDEF. 

Required,  to  inscribe  in  the  O 
a  A  equiangular  to  DEF. 

Analysis.  Suppose  the  problem 
solved,  and  that  ABC  is  the  re- 
quired A .  At  A  draw  the  tangent 
HAG. 

Then  Z  GAB  =  Z  C,  and  Z  HAC  =  Z  B,  (243) 

.  • .  the  Z  s  of  the  A  are  the  same  as  the  Z  s  GAB,  BAG, 
CAH. 

Hence,  the  following  construction: 

Cons,  At  any  pt.  A  on  the  ©ce,  draw  the  tangent  HAG. 

At  A  make  Z  GAB  =  Z  F,  aud  Z  HAC  =  Z  E.  (258) 

Join  BC.  Then  ABC  is  the  inscribed  A  required.     (100) 

55.  Given  two  sides  of  a  triangle  and  the  straight  line 
drawn  from  the  extremity  of  one  of  them  to  the  middle 
point  of  the  other,  to  construct  the  triangle. 

Construct  A  ABC  so  that  AB  =  1st  side,  BD  =  ^  the  2nd,  etc. 

56.  Given  two  sides  of  a  triangle  and  the  straight  line 
drawn  from  their  point  of  intersection  to  the  middle  point 
of  the  thh'd  side,  to  construct  the  triangle. 

Construct  the  a  ABE  so  that  AB  =  1st  side,  BE  =  3nd  side,  etc. 

57.  Given  the  base,  one  of  the  angles  at  the  base,  and 
the  sum  of  the  sides  of  a  triangle,  to  construct  the  trian- 
gle. 

58.  Given  the  base,  one  of  the  angles  at  the  base,  and 
the  difference  of  the  sides  of  a  triangle,  to  construct  the 
triangle. 


BOOK  IIAeXERQISES.     PROBLEMS.  131 

59.  Through  a  given  point  P  between  two  given  straight 
lines  AB,  AC^  draw  a  straight  line  which  shall  be  termi- 
nated by  AB,  AO,  and  bisected  in  P. 

60.  If  P  be  outside  the  lines  AB,  AC,  draw  PDE  meeting 
AB,  AC  in  D,  E  so  that  PD  equals  DE. 

61.  Find  points  D,  E  in  the  sides  AB,  AC  respectively 
of  the  triangle  ABC,  so  that  DE  may  be  parallel  to  BC  and 
equal  to  BD. 

62.  Draw  a  straight  line  DE  parallel  to  the  base  BC  of 
the  triangle  ABC  and  meeting  AB,  AC  in  D  and  E,  so  that 
DE  equals  the  sum  of  BD,  CE. 

63.  From  a  given  point  without  a  given  straight  line 
draw  a  straight  line  making  a  given  angle  with  this  given 
line. 

64.  AB,  AC  are  two  straight  lines,  AE  is  an  intermedi- 
ate straight  line.  Show  how  to  draw  the  straight  lines  which 
are  terminated  by  AB,  AC,  and  bisected  by  AE. 

Take  any  pt.  E  in  AE,  draw  EF  1|  to  AC  meeting  AB  in  F;  from  FB  cut  off  FG 
=  AF,  etc. 

65.  Construct  a  triangle,  having  given  a  median  and  the 
two  angles  into  which  the  angle  is  divided  by  that  median. 

QQ,  Construct  a  parallelogram,  having  given  (1)  two  di- 
agonals and  the  angle  between  them,  and  (2)  one  side,  one 
diagonal,  and  the  angle  between  the  diagonals. 

67.  From  a  given  point  0  draw  three  straight  lines  OA, 
OB,  OC  of  given  lengths  so  that  A,  B,  C  may  be  collinear, 
and  AB  equal  to  BC. 

Construct  a  a  GAD  having  the  sides  OA,  AD  =  the  two  outside  lengths,  and 
OD  double  the  middle,  etc. 

^^,  In  a  given  straight  line  find  a  point  whose  distance 
from  a  given  point  in  the  line  may  be  equal  to  its  distance 
from  another  given  straight  line. 

Let  P  be  the  given  pt.  in  the  given  line  XY,  AB  the  other  given  line.    Draw 
PC  J.  to  AB,  bisect  l  XPC,  etc. 

69.  Construct  a  triangle,  having  given  the  base,  the  ver- 
tical angle,  and  (1)  the  sum,  or  (2)  the  diiference  of  the 
sides. 


132  PLANE  GEOMETRY. 

Construct  a  right  triangle  having  given  : 

70.  The  hypotenuse  and  the  sum  of  the  sides. 

71.  The  hypotenuse  and  the  difference  of  the  sides. 

72.  The  hypotenuse  and  the  perpendicular  from  the 
right  angle  on  it. 

73.  Tiie  perimeter  and  an  angle. 

74.  Construct  an  isosceles  triangle,  having  the  vertical 
angle  equal  to  four  times  each  of  the  base  angles. 

75.  Construct  an  isosceles  triangle,  having  one-third  of 
each  angle  at  the  base  equal  to  half  the  vertical  angle. 

■  Circles. 

76.  Through  a  given  point  inside  a  circle  which  is  not 
the  centre,  draw  a  chord  which  is  bisected  at  that  point. 

77.  With  a  given  point  as  centre  describe  a  circle  which 
shall  intersect  a  given  circle  at  the  ends  of  a  diameter. 

78.  Describe  a  circle  which  shall  pass  through  a  given 
point  outside  or  inside  a  given  circle  and  touch  it  at  a  given 
point. 

79.  Describe  a  circle  with  a  given  centre  which  shall 
touch  a  given  circle.   How  many  such  circles  can  be  drawn  ? 

80.  Through  a  given  point  inside  a  given  circle  draw  two 
equal  chords  which  shall  contain  an  angle  equal  to  a  given 
angle. 

81.  Describe  a  circle  which  shall  touch  a  given  circle  at 
a  given  point,  and  also  touch  a  given  straight  line. 

Let  C  be  the  cent,  of  ©,  P  the  point.   Draw  the  diam.  A  BCD  i  to  given  line  XY, 
meeting  the  O  in  A;  join  PA,  PB,  and  produce  them  to  meet  XY  in  E,  F,  etc. 

82.  Describe  a  circle  passing  through  a  given  point  and 
touching  a  given  straight  line  at  a  given  point. 

83.  With  a  given  point  outside  a  given  circle  as  centre, 
describe  a  circle  which  shall  cut  the  given  circle  orthogo- 
nally (228). 

84.  Given  two  points  A,  B  on  a  circle,  and  a  fixed 
straight  line  through  A.     Draw  through  B  a  straight  liuQ 


BOOK  IL— EXERCISES.     CIRCLES.  133 

cutting  the  circle  in  C,  and  the  fixed  line  in  D,  so  that  AC 
shall  be  equal  to  CD. 

Let  the  fixed  line  meet  the  0  in  F;  bisect  the  arc  AB  in  E,  draw  EC  i  to  AF, 
etc. 

85.  Describe  a  circle  which  shall  touch  a  given  straight 
line  at  a  given  point,  and  pass  through  another  given  point 
not  in  the  line. 

86.  Construct  a  triangle,  having  given  the  base,  the  ver- 
tical angle,  and  the  median  drawn  from  the  vertical  angle. 

h  87.  ABC  is  a  given  straight  line:  find  a  point  P  such 
^hat  each  of  the  angles  APB,  BPC  may  be  equal  to  a  given 
angle. 

88.  Find  the  point  inside  a  given  triangle  at  which  the 
ides  subtend  equal  angles. 

89.  About  a  given  circle  to  describe  a  triangle  equiangu- 
lar to  a  given  triangle. 

On  two  sides  describe  segments  of  ©s  containing  an  Z  =  §  of  2  rt.  Zs,  etc. 

90.  If  the  escribed  circles  of  the  triangle  ABC  (272) 
touch  BC,  CA,  AB  externally  in  D,  E,  F  respectively,  prove 
that  BD  =  EA,  CD  =  AF,  CE  =  BF. 


Book  III. 

EATIO  AND  PEOPOETION.     SIMILAE 
riGUEES. 


Definitions. 

277.  Four  quantities  are  said  to  be  in  proportmi  when 
tlie  ratio  of  the  first  to  the  second  is  equal  to  the  ratio  of 
the  third  to  the  fourth. 

Thus,  if  g  =  5» 

then,  A,  B,  C,  D  are  said  to  be  in  proportion.    The  propor- 
tion is  written 

A  :  B  =  C  :  D, 

or  A  :  B  :  :  C  :  D, 

which  is  read  *^  A  is  to  B  as  C  is  to  D." 

Note.— The  first  form  is  preferable,  and  the  one  most  generally  used  in  the 
higher  mathematics;  but  the  second  form  is  the  more  usual  one  in  elementary 
works. 

Let  a  and  b  denote  the  numerical  measures  of  A  and  B 
(229),  and  c  and  d  the  numerical  measures  of  C  and  D. 
Then,  since  the  ratio  of  two  quantities  is  the  same  as  the 
ratio  of  their  numerical  measures  (230), 

,'.  A  :B  =  a  :b, 
and  C  :J)  =  c  :  d. 

Hence,  if  four  quantities  A,  B,  C,  D,  are  in  proportion, 
their  numerical  measures  a,  Z>,  6',  ^,  are  in  proportion;  that  is, 

a  :  b  =i  c  :  d^ 

134 


BOOK  IIL—BATIO  AND  PROPORTION.  135 

Conversely,  if  tlie  numerical  measures  of  four  quantities 
A,B,C,D,  are  in  proportion,  the  quantities  themselves  are 
in  proportion;  that  is, 

A  :  B  =  C  :  D, 

2uhen  A  and  B  are  quantities  of  one  kind,  and  C and  D  are 
quantities  of  one  hind,  though  the  latter  kind  may  be  dif- 
ferent from  the  former. 

That  is,  all  four  quantities  may  he  of  the  same  kind,  as, 
for  instance,  four  straight  lines,  four  surfaces,  four  angles, 
and  so  on;  but  the  quantities  in  each  pair  must  be  of  the 
same  kind. 

The  magnitudes  we  meet  with  in  Geometry  are  more 
often  incommensurahle  (232)  than  commensurable. 

The  preceding  reasoning  does  not  apply  directly  to  the 
case  in  which  two  quantities  are  incommensurable,  but  it 
may  be  extended  to  this  case. 

278.  To  find  the  greatest  common  measure  of  tioo  quan- 
tities. 

Let  there  be  two  quantities,  as,      j j ^      ^  ^ 

for  instance,  the  two  straight  lines  A  E  GB 
AB,  CD.  fe ^ 

Apply  the  smaller  CD  to  the 
greater  AB,  as  many  times  as  possible,  suppose  twice,  with 
a  remainder  EB. 

Apply  the  remainder  EB  to  CD  as  many  times  as  pos- 
sible, suppose  once,  with  a  remainder  FD. 

Apply  the  second  remainder  FD  to  EB  as  many  times  as 
possible,  suppose  once,  with  a  remainder  GB. 

Apply  the  third  remainder  GB  to  FD  as  many  times  as 
possible. 

This  process^ill  terminate  only  if  a  remainder  is  found 
which  is  contained  an  exact  number  of  times  in  the  pre- 
ceding one  ;  and  if  it  so  terminates  the  two  given  lines  are 
com  mensurable,  and  the  last  remainder  will  be  their  great- 
est common  measure. 


136  PLANE  GEOMETRY. 

Suppose,  for  example,  that  GrB  is  contained  exactly  twice 
ill  FD. 

Then  GB  is  contained  exactly  in  OF,  and  therefore  ex- 
actly in  CD,  and  therefore  exactly  in  AE,  and  therefore 
exactly  in  AB. 

Therefore  GB  is  a  common  measure  (231)  of  AB  and 
CD. 

And,  since  every  common  measure  of  AB  and  CD  must 
divide  AE,  it  must  divide  EB  or  CF,  and  therefore  FD, 
and  therefore  also  GB.  Hence,  the  common  measure  can- 
not be  greater  than  GB.  Therefore  GB  is  the  greatest 
common  measure  of  AB  and  CD. 

By  regarding  GB  as  the  measuring  unit,  the  values  of 
the  preceding  remainders  are  easily  found,  and  finally,  those 
of  the  given  lines,  from  which  their  numerical  ratio  is  ob- 
tained. 

Thus,  FD  =  2GB  ;     EB  =      FD  +    GB  =    3GB  ; 

CD=    EBH-FD=    2FD+    GB  =    5GB; 

AB  =  2CD  +  EB  =  10GB  +  3GB  =  13GB. 

Therefore  the  given  lines  AB  and  CD  are  numerically 
expressed  in  terms  of  the  unit  GB  by  the  numbers  13  and 
5  ;  and  their  ratio  is  ^-^-. 

279.  When  the  quantities  are  incommensurable  the 
above  process  never  terminates.  However  far  the  opera- 
tion is  continued,  we  never  find  a  remainder  which  is  con- 
tained an  exact  number  of  times  in  the  preceding  one. 
But  as  there  is  no  limit  to  the  number  of  parts  into  which 
AB  and  CD  may  be  divided,  we  may  obtain  a  remainder  as 
small  as  we  please,  one  that  is  less  than  any  assignable 
quantity. 

Hence,  although  no  fiyiite  numerator  and  denominator 
however  large  can  exactly  express  the  ratio  of  two  incom- 
mensurable quantities,  yet  by  properly  increasing  both 
numerator  and  denominator  we  may  obtain  a  ratio  as  nearly 


BOOK  III.— RATIO  AND  FEOrORTION.  187 

equal  as  we  please  to  the  required  ratio,  i.e.,  we  way  ohtain 
two  numbers  loliose  ratio  will  expres>>  the  ratio  of  two  in- 
commensurable quantities  to  any  required  degree  of  accu- 
racy. (232) 

Note.— We  therefore  conclude  that  ratio  in  Geometry  may  be  treated  in  the 
same  way  as  ratio  in  Arithmetic  and  Algebra.  Indeed,  the  algebraic  treatment 
is  the  easiest  and  the  simplest.  Euclid's  treatment  of  ratio  and  proportion  is 
now  pi-actically  disregarded.  While  his  reasoning  is  exquisite  and  rigorous,  it 
is  remote  from  our  practical  notions  on  the  subject.  For  these  reasons,  only 
simple  algebraic  proofs  of  the  propositions  in  proportion  will  be  given  in  this 
work.  The  student  will  perceive  that  the  propositions  do  not  introduce  new 
ideas,  but  merely  supply  proofs,  based  on  the  geometric  definition  of  propor- 
tion, of  results  already  familiar  in  the  study  of  Algebra. 

280.  Let  us  now  consider  the  numerical  proportion 
(277),  which  may  be  written  in  either  of  the  three  forms: 

a  :  b  =  c  :  d\     a  '.b\\  c  \  d\     —  =  —, 

b       d 

The  four  terms  of  the  two  equal  ratios  (277)  are  called 
the  ter^ns  of  the  proportion.  The  first  and  fourth  terms 
are  called  the  extremes^  and  the  second  and  third,  the 
means.  Thus,  in  the  above  proportion,  a  and  d  are  the 
extremes,  and  b  and  c  the  means. 

The  first  and  third  terms  are  called  the  antecedents,  and 
the  second  and  fourth  the  consequents.  Thus,  a  and  c  are 
the  antecedents,  and  b  and  d  the  consequents. 

The  fourth  term  is  called  a  fourth  iiroportional  to  the 
other  three.  T^hus,  in  the  above  proportion,  d  is  a  fourth 
proportional  to  a,  b,  and  c. 

In  the  proportion  «  :  Z>  =  J  :  c,  c  is  a  third  'proportional 
to  a  and  b,  and  Z^  is  a  mean jiroportional  between  a  and  c. 


138  PLANE  GEOMETRY. 


Proposition  1 . 

281.  If  four  quantities  are  in  proportion^  the  product 
of  the  extremes  is  equal  to  the  product  of  the  means. 

Hyp,  Let  a  '.h  —  c\  d. 

To  prove  ad  —  he. 

Proof.  By  {m),  |  =  ^. 

Multiplying  by  hd,        ad  =  be,  q.e.d. 

282.  Cor.  If  a:b  =  b:^c, 

,',  b'  =  ac,  (281) 

.-.  b  =  Vac, 
That  is,  the  mean  proportional  between  two  quantities  is 
equal  to  the  square  foot  of  their  product. 

Proposition  2. 

283.  Conversely,  if  the  product  of  two  quantities  be 
equal  to  the  product  of  two  others,  two  of  them  may  be  made 
the  extremes,  and  the  other  two  the  means,  of  a  proportion. 

Hyp.  Let  ad  =  be. 

To  prove  a  :  b  =  c  :  d. 

Proof,  Dividing  the  given  equation  by  bd, 

(I  _c  ^ 
b~l' 
that  is,  a  :  b  =  c  :  d,  q.e.d. 

In  a  similar  manner  it  may  be  shown  that  the  proportions 
a  '.  c  =  b  \  d, 
b  :  a  =  d  :  c, 
b  :  d  =  a  :  c, 
c  \  d  ^=^  a  \b,  etc., 
are  all  true  provided  that  ad  —  be. 


BOOK  UL-HATIO  AND  PUOPORTION.  139 

284.  Sen.  By  tlie  product  of  the  extremes  or  of  the 
meiiiis  of  a  proportion  is  meant  the  product  of  the  numerical 
measures,  of  those  quantities.  Hence,  the  product  of  tioo 
lines  will  be  often  used  for  brevity,  meaning  the  product  of 
the  numbers  luliich  represent  those  lines. 

Proposition  3. 

285.  If  fonr  quantities  are  in  proportioUy  they  are  in 
proportion  hy  inversion;  that  is,  the  second  term  is  to  the 
first  as  the  fourth  term  is  to  the  third. 

Hyp,  Let  a  -.h  —  c  \  d. 

To  prove  h  \  a  =  d  \  c. 

Proof  |  =  |.  (277) 

•   -^  •    b-^  '  d' 


that  is. 


b_d 

a       c  ' 

,'.  b  :  a  =  d  :  c,  q.e.d. 

Proposition  4. 

286.  If  four  quantities  are  in  proportion,  they  are  in 
proportion  by  alternation;  that  is,  the  first  term  is  to  the 
third  as  the  second  term  is  to  the  fourth. 

Hyp.  Let  a  :  b  =  c  :  d. 

To  prove  a  :  c  =  b  :  d. 

Proof  Since  a  :  b  =  c  :  d, 

, ' ,  ad  =  be.  (281) 

.'.  a:c:=b:d,  (283) 

Q.E.D. 


140  PLANE  GEOMETRY. 


Proposition  5. 

287.  If  four  quantities  arc  in  proportion,  they  are  in 
pro^jortion  by  composition;  that  is,  the  sum  of  the  first  and 
second  is  to  the  second  as  the  sum  of  the  third  and  fourth 
is  to  the  fourth. 

Hyp.  Let  a  :  b  =  c  :  d. 

To  prove  a-{-b  :b  =  c  -{-  d  :  d. 

Proof.  ^  =  -^.  (277) 


a 
1~ 

c 

'7v 

■1+- 

^J+>> 

a-{-b 

c-^-d 

that  is,  _  . 

0  d 

.'.  a  +  b  :  b  =  c  -\-  d  :  d.  Q.E.D. 

Similarly,  a  +  b  :  a  =  c  ^  d  :  c. 

Proposition  6. 

288.  If  four  quantities  are  in  proportion,  they  are  in 
proportion  by  division;  that  is,  the  difference  of  the  first 
and  second  is  to  the  second  as  the  difference  of  the  third 
and  fourth  is  to  the  fourth. 

Hyp.  Let  a  -.b  =  c  :  d. 

To  prove  a  —  b  :  b  =  c  —  d  :  d, 

(277) 


Proof. 

a       c 
b  ~d' 

•   •  b      ^-d      ^' 

that  is. 

a  —  b       c  ^rz^ 
b      ~      d     ' 

.' .  a  —  b  :  b  —  c  —  d  :  d. 

Similarly, 

a  —  b  \  a  —  c  —  d  \  c. 

Q.E.D. 


BOOK  III.— RATIO  AND  PROPORTION.  141 


Proposition  7. 

289.  Jf  four  qiiantities  are  in  proportion,  they  are  in 
proportion  by  composition  and  division;  that  is,  the  sum  of 
the  first  and  second  is  to  their  difference  as  the  sum  of  the 
third  and  fourth  is  to  their  difference. 

Hyp.  Let  a:b  —  c\d. 

To  prove       a  -\-  b:  a  —  b  =^  c  -\-  d:  c  —  d. 

Proof.  By  (287),  — 

and  by  (288) 
by  division, 

d.  Q.E.D. 

Proposition  8. 

290.  The  products  of  the  corresponding  terms  of  two  or 
more  proportions  are  projjortional. 

Hyp.  Let     a:  b  =  c  :  d,  and  e:  f  =  g  \h. 
To  prove  ae :  bf  =  eg :  dh. 

r)      y  a       c        ^  e       g 

Proof  -7  =  -J  J  and  -.  =  f . 

•^  h      d'  f       h 

ae  _  eg 

■'■  ¥"//'■ 

.*.  ae:  of  =  eg:  dh.  q.e.d. 

291.  A  greater  quantity  is  said  to  be  a  inultiple  of  a 
less,  when  the  greater  contains  the  less  an  exact  number  of 
times. 

BquimuUiples  of  two  quantities  are  quantities  which 
contain  them  the  same  number  of  times.  Thus,  ma  and 
mb  are  equimultiples  of  a  and  b. 


V^o<;, 

h     - 

'      d     ' 

a  —  b 

b      ~ 

c  —  d, 
■      d     ' 

a-\-b 
a  —  b 

c^d 
'  c-d' 

.\a  +  b 

:a-b: 

=  c  +  d: 

c 

149  PLANE  OEOMBTRY. 


Proposition  9. 

292.  Wlien  four  qvantities  are  in  proportion,  if  the 
iird  and  second  he  multiplied^  or  divided,  hy  any  quantity, 
as  also  the  third  and  fourth,  the  resulting  qua^itities  will 
he  in  proportion. 


Hyp,  Let 

a:l)  =  c:d. 

To  prove 

ma  :mh=  nc\  nd. 

Proof, 

a       c 

Multiply  both  terms  of  the  first  fraction  by  m,  and  both 
terms  of  the  second  by  n. 


Then 


Similarly, 


w,a  _  nc 
mh  ~  nd 

ma  :  mh  =  nc  :  nd,  q.e.d. 

a     h  _  c    d 
m '  m      n'  n 


293.  ScH.  Either  m  or  n  may  be  unity. 

In  a  similar  manner  it  maybe  shown  that  if  the  first  and 
third  terms  he  multiplied,  or  divided,  hy  any  quantity,  and 
also  the  second  and  fourth,  the  resulting  quantities  will  he 
in  proportion. 


294.  Cor.  Since 


ma  _  a 
mh       h 


a  _a 
h~h 

.'.  ma  :  mh  =  a:  h. 


That  is,  equimtdtiples  of  two  quantities  are  in  the  same 
ratio  as  the  quantities  tliemselves. 


BOOK  ilL-PROPORTIONAL  LINES.  143 

Proposition  lO. 

295.  If  four  quantities  are  in  proportion,  their  like 
poivers,  or  roots,  are  in  jjroportion. 


Hyp.     Let 
To  prove 

and 

a',h  =  c:  d. 
«»  :  Z.»  =  c"  :  (?«, 

11        11 
a^  \  !)■»■  — c^\  d\ 

Proof 

a       c 
b~d' 

Kaising  to  the 
Extracting  the 

n^^  power, 

n*^  root, 

1        1 

J«  =  c"^ !  d\ 


111 
J»  =  6*»  :  d"" 


Q.E.D. 

Proposition  11. 

296.  If  any  number  of  quantities  are  in  proportion, 
any  antecedent  is  to  its  consequent,  as  the  sum  of  any  num- 
ber of  the  antecedents  is  to  the  sum  of  the  corresponding 
consequents. 

Hyp.        Let  a  :  b  =  c  :  d  =  e  :  f  ^  etc. 

To  jjrove 

a\b  =  c\d—  etc.  —  a -\- c -\- e -\-  etc.  \b-{-d  +/+  etc. 

Proof.  __ab  =  ba^ 

ad  =  be,  and  af=  be,  etc.,  etc.  (281) 

Adding,  a{b  +  d  +/-f  <^tc.)  =  b{a  -\-  c  -{-  e  -\-  etc.) 
,\a:b  =  c  :d  =  etG.  =  a  +  c  +  e-^etG.  :b  +  d+f +etG.  (283) 

Q.E.D. 

Proportional  Lines. 

297.  Def.  Two  straight  lines  are  said  to  be  cut  propor- 
tionally when  the  parts  of  one  line  are  in  the  same  i-atio  as 
the  corresponding  parts  of  the  other  line. 

Thus,  AB  and  CD  are  cut  pro-  P  ^ 

portionally  at  P  and  Q  if  A 


AP  :  PB  =  CQ  :  QD.  k ^ 


144 


PLANE  GEOMETRY. 


Proposition  1 2.    Theorem.  . 

298.  A  straight  line  parallel  to  one  side  of  a  triangle 
divides  the  other  tivo  sides  proportionallg. 
Hy2J.  Let  DE  be  II  to  BC  in  the  A  ABC.  A, 

To  prove    AD  :  DB  =  AE  :  EC.  '^' 

Case  I.  Wlieu  AD  and  DB  are  com- 
mensurable. 

D/ ^-^'^ 

Proof.  Take  AH,  any  common  meas- 
ure of  AD  and  DB,  and  suppose  it  to 
be  contained  4  times  in  AD  and  3  times 
inDB.  ^  ^ 

Then  ^5  =  |.  (1) 

Through  the  several  pts.  of  division  of  AD,  DB  draw  ||  s 
to  BC.  They  will  divide  AE  into  4  equal  parts  and  EC 
into  3  equal  parts. 

Jf  \\  s  iniei'cepi  equal  lengths  on  any  transversal,  tliey  int€rccpt  equal 
lengths  on  evei'y  transtei'sal  (152). 

•   •  EC  -  3  •  ^^^ 

Therefore,  from  (1)  and  (2), 

DB~  EC  ^         ^ 

.  • .  AD  :  DB  =  AE  :  EC. 

Case  II.  When  AD  and  DB  are  incom-  Aa 

inensnrable. 

Proof.  In  this  case  we  know  (232)  that 
we  may  always  find  a  line  AG  as  nearly 

equal  as  ive  please  to  AD,  and  such  that       G/ \H 

AG  and  GB  are  commensurable. 

Draw  GHllto  BC  ;  then 

AG      AH  ,p       Tx    ^  *         ^ 


BOOK  IIL-PROPORTIONAL  LINES. 


145 


(233) 
Q.E.D 


Now,  these  two  ratios  being  always  equal  while  the  com- 
mon measure  is  indefinitely  diminished,  they  will  be  equal 
when  G  moves  up  to  and  as  nearly  as  we  please  coincides 
with  D. 

ADAE 
•**DB~EO* 

299.  Cor.  1.     By  composition  (287),  we  have 

AD  +  J3B  :  AD  =  AE  +  EC  :  AE, 
or  AB  :  AD  =  AC  :  AE. 

Also,  AB  :  DB  =  AC  :  EC, 

and  AB:AC  =  DB:EC. 

300.  Cor.  2.  If  Uvo  straight  lines 
AB,  CD  are  cut  hy  any  mimber  of 
2)arallels,  AC,  EF,  GH,  BD,  the  cor- 
responding intercepts  are  proportional. 

For,  let  AB  and  CD  meet  at  0. 
Then,  by  (299), 


OE 
OF 


AE 
~CF 


EG 
"FH* 


(288) 
(286) 

t" 

e/ 

\f 

J 

\„ 

1               \ 

B 


Similarly, 


AE  :  CF  =  EG  :  FH. 
EG  :  FH  =  GB  :  HD. 


EXERCISES. 

1.  From  a  point  E  in  the  common  base  AB  of  two  tri- 
angles ACB,  ADB,  straight  lines  are  drawn  parallel  to  AC, 
x\D,  meeting  BC,  BD  at  F,  G  :  show  that  EG  is  parallel  to 
CD. 

2.  In  a  triangle  ABC  the  straight  line  DEF  meets  the 
sides  BC,  CA,  AB  at  the  points  D,  E,  F  respectively,  and 
it  makes  equal  angles  with  AB  and  AC  :  prove  that 

BD  :  CD  =  BF  :  CE. 


146  PLANE  GEOMETRY. 

Proposition  13.    Theorem. 

301.  Conversely,  if  a  straight  line  divides  two  sides  of 
a  triangle  proportionally,  it  is  parallel  to  the  third  side. 

Hyp,     Let  DE  cut  AB,  AC  in  a 

the  A  ABC  so  that       -p=-  =  -j-=,  /        \ 

AD      AE  /  \   / 

To  prove  DE  ||  to  BC.  9^ "  V 

Proof      If  DE  is  not  ||  to  BC,    ^/_ \^ 

draw  DE'  ||  to  BC.  B  0 

Then, 

But  T^  =  Ti-  (Hyp-) 

(Ax.  1) 

.•.AE'  =  AE, 
which  is  impossible  unless  DE^  coincides  with  DE. 

.-.  DEis  II  toBC.  Q.E.D. 

302.  Def.    When  a  finite  straight  line,  as  AB,  is  cut  at 
a  point  X  between  A  and  B,  it  is 

said  to  be  divided  internally  at  X,      l 1 — ^— — jY 

and  the  two  parts  AX  and  BX  are 

called  segments.  But  if  the  straight  line  AB  is  produced, 
and  cut  at  a  point  Y  beyond  AB,  it  is  said  to  be  divided 
externally  at  Y,  and  the  parts  AY  and  BY  are  called  seg- 
ments. The  given  line  is  the  sum  of  two  internal  seg- 
ments, or  the  difference  of  two  external  segments. 


AB 

AC 

AD" 

AE' 

AB 

AC 

AD 

~AE- 

AC 

AC 

AE' 

"AE" 

BOOK  III— PROPORTIONAL  LINES.  147 


Proposition  14.    Theorem. 

303.  The  hisector  of  an  angle  of  a  triangle  divides  the 
opposite  side  into  segments  proportional  to  the  adjacent 
sides. 

Hyp.  Let  AD  bisect  the  Z  A  of  the 
A  ABC. 

To  prove    DB  :  DC  =  AB  :  AC. 

Proof  Draw  CE  !|  to  DA,  meeting 
BA  produced  in  E. 

Since  DA  is  H  to  CE,    (Cons.) 

.-.  Z  BAD  =  Z  AEC, 
being  ext. -int.  Zsof\\  lines  {11), 

and  .     Z  CAD  =  Z  ACE, 

being  alt. -int.  As  of  \\  lines  (72). 

But  Z  BAD  =  Z  CAD.  (Hyp.) 

.-.  Z  AEC  =  I  ACE.  (Ax.  1) 

.-.  AE  =  AC, 

being  opp.  equal  As  (114). 

Because     DA  and  CE  are  1| , 

.-.  DB  :  DC  =  AB  :  AE.  (298) 

But                      AE  =  AC.  (Just  proved) 

.-.  DB  :  DC  =  AB  :  AC.  q.e.d. 

Cor.  Conversely,  if  AD  divides  BO  into  tioo  segments 
that  are  proportional  to  the  adjacent  sides,  it  bisects  the 
angle  BA  C. 


148 


PLANE  GEOMETRY. 


Proposition    1 5.    Theorem. 

304.  Tlie  bisector  of  cm  exterior  angle  of  a  triangle 
divides  the  opposite  side  externally  into  segments  2^ropo7'- 
tio7ial  to  the  adjacent  sides. 

Hyp,  Let  AD  bisect  the   ext. 
Z  CAH  of  the  A  ABC. 

To  prove  DB  :  DC  =  AB  :  AC. 

Proof,  Draw     CE    ||    to    DA 
meeting  BA  at  E. 

Since  DA  is  ||  to  CE,    (Cons.) 


(77] 


and 
But 


.-.  Z  AEC  =  Z  DAH,  c.^r^\J^  U 

Z  ACE  =  Z  DAC.   oulA.  L  (72) 

Z  DAH  =  Z  DAC,    K^cH,/   (Hyp.) 


.-.  Z  AEC  =  Z  ACE.  (Ax.  1) 

.-.  AE  ==  AC.  (144) 
Because  DA  and  CE  are  II , 

.-.  DB  :  DC  =  AB  :  AE.  (299) 

But                              AE  =  AC.  (Just  proved) 

.-.  DB  :  DC  =  AB  :  AC.  q.e.d. 

305.  Cor.  If  A F  be  the  bisector  of  the  interior  angle 
BAC,  we  have,  from  (303)  and  (304), 

BE  :  EC  =  BD  :  CD. 

306.  Def.  "When  a  straight  line  is  divided  internally 
and  externally  into  segments  having  the  same  ratio,  it  is 
said  to  be  divided  Imrmonically,     Therefore  (305), 

The  bisectors  of  an  interior  and  exterior  angle  at  the 
vertex  of  a  triangle  divide  the  opposite  side  harmonically. 


BOOK  IIL-SIMILAU  FIGURES. 


149 


EXERCISES. 

1.  li  a  ',  h  =  e  '.  fi  and  c  :  d  =  e  \  f,  show  that  a  :  J  = 
c :  d, 

2.  \i  a-.l  —  l',  c,  show  that  a\  c  —  o}  '.h^, 

3.  A,  B,  C,  D  are  four  points  on  a  straight  line,  E  is  any 
point  outside  the  line.  If  the  parallelograms  AEBF, 
OEDG  be  completed,  show  that  EG  will  be  parallel  to  AD. 

Similar  Figures. 

307.  Def.  Similar  figures  are  those  which  are  mutually 
equiangular  (144),  and  have  their  homologous  sides  pro- 
portional. 

The  homologous  sides  are  those  which  are  adjacent  to 
the  equal  angles.  (145) 


Thus,  the  figures  ABCD,  EFGH  are  similar  if 
ZA=  ZE,  ZB=  ZF,  ZC=  ZG,  etc.; 


and 


AB_  BC       op      DA 
EF  ~  EG  ~  GH  ~  HE ' 


The  sides  AB  and  EF  are  homologous,  since  they  are 
adjacent  to  the  equal  angles  A  and  B,  and  E  and  F,  respec- 
tively; also  the  sides  BC  and  EG  are  homologous,  etc.  The 
diagonals  AC,  EG  are  homologous. 

308.  The  constant  ratio  of  any  two  homologous  sides 
of  similar  figures  is  called  the  ratio  of  similitude  of  the 
figures. 


150  PLANE  GEOMETBT. 


Proposition   1 6.    Theorem. 

309.  Triangles  luhicli  are  r}iutuaUy  equiangular  are 
similar. 

Hy2h  In  the  As  ABC,  A'B'C,  A. 

let    Z  A  =  Z  A',    Z  B  =  Z  B',  A 

zc=  zc\  /   \  ^ 

To  prove     A  ABC    similar    to       qA \g         /  \ 

aA'B'C  /  A     /     \ 

Proof.    Apply   the    A  A'B'C     ^ ^    ^ )^'_ 

to  the  A  ABC  so  that  the  pt.  A' 

coincides  with  A,  and  A'B'  falls  on  AB.     Let  B'  fall  at 

D. 

Then,  since  Z  A'  =  Z  A,  (Hyp.) 

.  • .  A'C  falls  on  AC. 
Let  C  fall  at  E;  then  B'C  falls  on  DE. 

Since  ZADE=zABC,  (Hyp.) 

.-.  DEislltbBC.  (78) 

.-.  AB:  AD  =  AC:  AE,      /  (299) 

or  AB  :  A'B'  =  AC  :  A'C. 

In  the  same  way,  by  applying  the  A  A'B'C  to  the 
A  ABC  so  that  the  Z  s  at  B'  and  B  coincide,  we  may 
prove  that 

AB  :  A'B'  =  BC  :  B'C. 
Combining  these  two  proportions,  we  have 

AB  :  A'B'  =  AC  :  A'C  =  BC  :  B'C. 

.  • .  the  two  A  s  are  similar.  (307) 

Q.E.D. 

310.  Cor.  1.  Ttuo  triangles  are  similar  when  two 
angles  of  the  one  are  eqnal  respectively  to  tivo  angles  of  the 
other.  '  (97) 

311.  Cor.  2.  A  triangle  is  similar  to  any  triangle  cut 
off  hy  a  li7ie  parallel  to  one  of  its  sides, 

312.  ScH.  In  similar  triangles  the  homologous  sides  lie 
opposite  the  equal  angles, 


BOOK  IIL— SIMILAR  FIGURES, 


151 


Proposition  1  7.    Theorem. 

313.  Triangles  which  have  their  homologous  sides  pro- 
portional are  similar. 

Hyp.  In  the   As  ABO,  A'B'C, 

AB  _  AC  _  BC 
let         ^,-g,  -  ^,^,  -  ^,^, . 

To  prove    Z  A  =  Z  A',  Z  B  = 
ZB',  etc.  C\  '        ' 

Proof.      On    AB    take    AD  ==    , 
A'B',  and  draw  DE  ||  to  BO. 

Then,  because  DE  is  I|  to  BC, 

. • .  the  AS  ABC,  ADE  are  similar. 

ABACBC 
•'•  AD~AE~DE* 
But  A'B'  =  AD. 

AC 


AB 
AD 


A'C' 


BO 
B'C 


(311) 

(307) 

(Cons.) 

(Hyp.) 


Since  the  first  ratio  in  each  of  these  proportions  is  the 
same,  the  second  and  third  ratios  in  eacli  are  equal  respec- 
tively ;  that  is. 


AC 
AE 


AC 


^    BO       ^^ 
and   :f^.  =  :=^^, 


BC 
B'C 


A'O'^    —  J)  J. 

Since  the  numerators  are  the  same, 

.  • .  AE  =  A'C,  and  DE  =  B'C. 

.-.  A  A'B'0'=  A  ADE, 
having  the  three  aides  equal  each  to  each  (108). 
But  A  ADE  is  similar  to  the  A  ABC. 

.  • .  A  A'B'C'  is  similar  to  the  A  ABC.         q.e.d. 

EXERCISE. 

If  the  sides  of  the  triangle  in  Prop.  14  are  AB  =  8, 
BO  =  12,  and  AC  =  10,  find  the  lengths  of  the  segments 
BD  and  CD. 


152 


PLANE  GEOMETRY, 


Proposition  1  8.    Theorem. 

314.  Two  triangles  which  have  an  angle  of  the  one 
equal  to  an  angle  of  the  other,  and  the  sides  about  these 
angles  proportional,  are  similar. 

Hyp.  Ill  the  A  s  ABC,  A'B'C,  let 

AB  _  AC 
A'C 


Z  A  =  Z  A',  and 


A'B' 

A  ABC    similar 


to 


A'B', 


(311) 


To  prove 
A  A'B'C. 

Proof.  On  AB  take  AD 
and  draw  DE  ||  to  BC. 
Because  DE  is  |I  to  BC, 

.• .  the  AS  ABC,  ADE  are  similar. 
.-.AB:  AD  =  AC:  AE, 
homologous  sides  of  similar  as  are  jn'oportional  (307). 
But  A'B'  =  AD.  (Cons.) 

.• .  AB  :  AD  =  AC  :  A'C.  (Hyp.) 

.• .  A'C  =  AE.  (Comparing  the  two  proportions) 
.-.  A  A'B'C  =  A  ADE, 
having  two  sides  and  ilie  included  /.  equal,  eacJi  to  each  (104). 

.  • .  A  A'B'C  is  similar  to  the  A  ABC.  q.e.d. 
315.  ScH.  From  the  definition  (307),  it  is  seen  that 
two  conditions  are  necessary  that  polygons  may  be  similar : 
(1)  they  must  be  mutually  equiangular,  and  (2)  their 
homologous  sides  must  be  proportional.  In  the  case  of 
triangles  we  learn  from  Props.  16  and  17  that  each  of  these 
conditions  follows  from  the  other;  so  that  one  condition  is 
sufficient  to  establish  the  similarity  of  triangles. 

This,  however,  is  not  necessarily  the  case  with  polygons 
of  more  than  three  sides;  for  even  with  quadrilaterals,  the 
angles  can  be  changed  without  altering  the  sides,  or  the 
proportionality  of  the  sides  can  be  changed  without  alter- 
ing the  angles. 


BOOK  IIL—SIMILAB  FIGUIiES. 


im 


Proposition  1 9.    Theorem. 


( 


316.  Tioo  triangles  tuhich  have  their  sides  parallel  or 
perpendicular,  each  to  each,  are  similar. 

Hyp,  In  the  As  ABC,  A'B'C,  ^  ^, 

let  AB,  AC,  BC  be  ||  or  J_,  respec- 
tively, to  A'B',  A'C,  B'C. 

To  prove  A  ABC  similar  to 
A  A'B'C.  B 

Proof.  Since  the  sides  are  I|  or 
J_  each  to  each,  the  included  Z  s 
are  equal  or  supplementary  (80, 
83). 

.  • .  three  hypotheses  may  be  made : . 

(1)  A  +  A'  =  2rt.Zs,B  +  B'  =  2rt.Zs,C  +  C'  =  2rt.Zs. 

(2)  A  =  A',         B  +  B'  =  2  rt.  Zs,  C  +  C  =  2rt.Zs. 


(3) 


B  =  B', 


C  =  C'. 


The  first  and  second  hypotheses  are  inadmissible,  since 
the  sum  of  the  Z  s  of  the  two  A  s  cannot  exceed  4  rt.  Z  s 
(97).     Therefore  the  third  is  the  only  admissible  hypothesis. 


the  A  s  ABC,  A'B'C  are  similar, 
being  mutually  equiangular  (309), 


Q.E.D. 


317.  ScH.  The  homologous  sides  of  the  two  triangles 
are  either  the  parallel  or  the  perpendicular  sides. 


EXERCISE. 


If  the  sides  of  the  triangle  in  Prop.  15  are  AB  =  16, 
BC  =  10,  and  AC  —  8,  find  the  lengths  of  the  segments 
BD  and  CD. 


154 


PLANE  GEOMETUY, 


Proposition  20.    Theorem. 

318.  If  in  any  triangle  a  2J(irallel  he  draion  to  the  base, 
all  lines  from  the  vertex  will  divide  the  base  and  its  parallel 
proportio7ially. 

Hyp.  Let  ABO  be  a  A,  WQ'  \\  to 
BC,  and  AD,  AE  two  lines  intersect- 
ing B'C  at  D'E'. 

BD         DE         EC 
To  prove      ^_  =  j^_  =-g^. 

Proof  Since  B'C  is  ||  to  BC,  the 
AS  AB'D',  ABD  are  similar;  also  the 
AS  AD'E'  and  ADE;  and  the  As  AE'C  and  AEC.    (311) 

AD        BD         ,  AD        DE 


__BD^       ^i^- 

•*•  AD' ""  B'D''        AD'  ~  D'E' ' 

their  Jwmologous  sides  being  proportional  (307). 

BD    _  DE 
•*' B'D"'~"D'E'' 

In  the  same  way  it  may  be  shown  that 

DE         EC 


(Ax.  1) 


D'E' 
BD 


E'C 
DE 


B'D'  ~  D'E' 


EC 
E'C 


Q.E.D. 


319.  Cor.  If  BD  =  DE  =  EC,  we  shall  have  B'D' 
=  D'E'  =  E'C.  Therefore,  if  the  lines  from  the  vertex 
divide  the  base  into  equal  parts,  they  will  also  divide  the 
parallel  into  equal  parts, 

EXERCISES. 

1.  If  BD  =  8,  B'D'  =  6,  DE  =  10,  and  EC  =  3,  find 
the  lengths  of  D'E'  and  E'C. 

2.  If  AD  =  20,  AD'  =  16,  AE  =  15,  BD  =  7,  and 
DE  =  5,  find  B'D',  D'E',  and  AE', 


BOOK  III.— SIMILAR  FIGURES. 


155 


Proposition  2 1 .    Theorem. 

320.  If  two  2yolygo7is  are  composed  of  the  same  numler 
of  triangles  similar  each  to  each,  and  similarly  placed,  the 
jjolygons  are  similar. 

Hyp,  Let  the  as  ABC,  ACD,  ADE 
of  the  polygon  ABODE  be  similar  re- 
spectively to  the  AsA'B'C,  A'C'D', 
A'D'E'  of  the  polygon  A'B'C'D'E',  and 
similarly  placed. 

To  prove  the  polygons  are  similar. 

Proof.  Since  the  homologous  Zs  of 
similar  A  s  are  equal,  (307) 

.-.     ZB=ZB'. 


Also, 
and 


Z  AOB 

Z 


(307) 


A(Ji5  =  Z  A'U'B',  ) 
ACD  =  Z  A'C'D'.  [ 

Adding,  Z  BCD  =  z  B'C'D'. 

In  like  manner,  Z  CDE  =  z  C'D'E' ;  etc. 


.*.  the  polygons  are  mutually  equiangular. 
Since  similar  as  have  their  homologous  sides  propor- 
tional, (307) 


AB 


BC 


AC 


AC 


A'B'  ~  B'C  ~  A'C"  ^^^  A'C 


CD 
"CD' 


AB 


BC 


A'B'      B'C 


CD 
CD 


7  =  etc. 


(Ax.  1) 


.  • .  the  homologous  sides  of  the  polygons  are  proportional. 
.  • .  the  polygons  are  similar, 

heing  mutually  equiangular  and  hamng  tJieir  Iwmologous  sviies 

proportional  (307).  Q.E.Q, 


156 


PLANE  GEOMETRY. 


Proposition  22.    Theorem. 

321.  Conversely,  two  similar  polygons  may  he  divided 
into  the  same  number  of  triangles,  similar  each  to  each,  and 
similarly  placed. 

Hyp.  Let  ABCDE,  A'B'C'D'E'  be 
two  similar  polygons  divided  into  As  by 
the  diagonals  AC,  AD,  A'C,  A'D'  drawn 
from  the  homologous  Z  s  A  and  A'. 

To  prove  As  ABC,  ACD,  ADE  sim 
ilar  respectively  to  As  A'B'C,  A'C'D', 
A'D'E'. 

Proof.  Since  the  polygons  are  similar, 

^,       ^   AB         BC 
...  ZB=zB,and^^  =  g^. 

.-.A  ABC  is  similar  to  A  A'B'C.  (314) 

.-.   Z  ACB=  Z  A'C'B'.  (307) 

But  Z  BCD  =  Z  B'C'D'.  (307) 

.  • .  remaining  Z  ACD  =  remaining  Z  A'C'D'. 
.,  BC    _  AC         ,   BC    _  CD  ,_^, 

^iso,        ]V(7  ~  K'Q"  ^     WW  ~  cTy '         (^^ ' ) 

.-.  AC  :  A'C  =  CD  :  CD'.  (Ax.  1) 

.  • .   A  ACD  is  similar  to  A  A'C'D'.  (314) 

In  the  same  way  it  may  be  shown  that  A  s  ADE,  A'D'E' 
are  similar.  q.e.d. 

Cor.  The  homologous  diagonals  of  two  similar  j^olygons 
are  j^ropor'tional  to  the  ho^nologous  sides. 


BOOK  III.—SIMILAH  FIGUUE8.  157 


Proposition  23.    Theorem. 

322.  Tlie  perimeters  of  tioo  similar  polygons  are  to  each 
other  as  any  tivo  homologous  sides. 

Hyp,  Let  ABCDE,  A'B'C'D'E'  be  two 
similar  polygons;  denote  their  perimeters 
by  P  and  P'. 

To  prove    P  :  P' =  AB  :  A'B'. 

Proof.  Since  the  polygons  are  similar, 
AB         BO  CD  .      ^_,. 

AB  4-  BC  +  CD  +  etc.     _   AB 
• '  •  A'B'  +  B'C  +  CD'  +  etc.  ~  A'B'* 

,-.P:P'  =  AB:  A'B'. 

EXERCISES. 

1.  From  the  ends  of  a  side' of  a  triangle  any  two  straight 
lines  are  drawn  to  meet  the  other  sides  in  P,  Q ;  also  from 
the  same  ends  two  lines  parallel  to  the  former  are  drawn  to 
meet  the  sides  produced  in  P',  Q':  show  that  PQis  parallel 
to  P'Q'. 

Let  ABC  be  the  a  ;  BP,  CQ  the  lines.    AC  :  AP  =  AQ'  :  AB,  etc. 

2.  ABC  is  a  triangle,  AD  any  line  drawn  from  A  to  a 
point  D  in  BO  ;  a  line  is  drawn  from  B  bisecting  AD  in  E 
and  cutting  AC  in  F :  prove  that  BF  is  to  BE  as  2  OF  is 
to  AC. 

Draw  EG  ||  to  AC  meeting  BC  in  G.    DE  =  EA,  .*.  AC  =  3  EG,  etc. 

3.  If  in  Prop.  20,  AD  =  24,  AD'  =  20,  AE  =  IG,  BD  ^ 
8,  and  DE  =  G,  jfind  B'D',  D'E',  and  AE', 


^ 


vo 


158 


PLANE  GEOMETRY. 


Numerical  Eelations  between-  the  Diffeeent 
Parts  of  a  Triangle. 


Proposition  24.    Theorem. 

323.  In  a  right  triafigle,  if  a  perpendicwlar  he  drawn 
from  the  right  angle  to  the  hypotenuse : 

(1)  The  two  triangles' on  each  side  of  it  are  similar  to  the 
tvhole  triangle  and  'to  each  other, 

(2)  Tlie  perpendicular  is  a  mean  proportional  between  the 
segments  of  the  hypotenuse, 

(3)  Each  side  ahoiitthe  right  angle  is  a  mean  proportional 
between  the  hypoten'nse  and  the  adjacent  segment. 

Hyp.  Let  ABO  be  a  rt.  A ,  and  AD 
the  _L  from  the  rt.  Z  A  to  the  hypote- 
nuse BO. 

(1)  To  prove  i\iQ  As  DAB,  DAO, 
and  ABO  similar. 

Proof     In  the    rt.   As   DAB, 

AOB,    the  acute  Z  B  is  common. 

.  • .  the  A  s  DAB,  AOB  are  similar.  (309) 

Also,  in  the  rt.  A  s  DAO,  ABO, 

the  acute  Z  0  is  common. 

.  • .  the  A  s  DAO,  ABO  are  similar.  (309) 

.  • .  the  A  s  DAB,  DAO  are  similar  to  each  other,  as 

they  are  both  similar  to  ABO.  q.e.d. 

(2)  To  prove        BD  :  AD  =  AD  :  OD. 

Proof     Since  the  A  s  DAB,  DAO  are  similar, 

.-.BD:  AD  =  AD:OD.  (307) 

(3)  To  prove        BO  :  AB  =  AB  :  BD. 

Proof.     Since  the  A  s  BAO,  BAD  are  similar, 


BOOK  III— NUMERICAL  MEASURES  IN  A  TRIANGLE.  159 

».-.  BC  :  AB  =  AB  :  BD.  (307) 

Iso,  since  the  A  s  CAB,  CAD  are  similar, 
.-.  BC  :  AC  =  AC  :  CD.  q.e.d. 

324.  Cor.  1.  If  the  lines  of  the  figure  are  expressed  in  \ 
Qumbers  by  means  of  their  numerical  measures  (277),  we 
have  from  the  above  three  proportions,  by  means  of  (281),  / 

/  AD'  =  BD  X  CD,  n' 

AB'  =  BC  X  BD,  V 

AC'  =  BC  X  CD. 
By  division,  the  last  two  equations  give 

AB'_  BD 

AC'  ~  CD-  J 

Hence,  tlie  squares  of  the  sides  about  the  right  angle  are 
vroportio7ial  to  the  adjacent  segments  of  the  hyjjotenuse. 

325.  Cor.  2.  Since  an  angle  inscribed 
in  a  semicircle  is  a  right  angle  (240), 
therefore:  (323) 

(1)  The  perpendicular  fro7n  any  point 
in  the  circumference  of  a  circle  to  a  diam-  "  ^ 
ter  is  a  mea7i  proportio7ial  betiveen  the  segments  of  the 
diameter. 

(2)  The  chord  from  the  point  to  either  extremity  of  the 
diameter  is  a  mea7i  proportional  between  the  diameter  and 
the  adjacent  segment, 

326.  Def.  Tlie  projection  of  a  point 
A  upon  an  indefinite  straight  line  XY 
is  the  foot  C  of  the  perpendicular  let 
fall  from  the  point  to  the  line. 

X      C  D      Y 

The  projection  of  a  ji)iite  straight  line 
AB  upon  the  line  XY  is  the  part  of  XY  between  the  per- 
pendiculars dropped  from  the  ends  of  the  line  AB. 

Thus,  CD  is  the  projection  of  AB  upon  the  line  XY. 


160 


PLANE  GEOMETRY. 


Proposition  25.    Theorem. 

327.  21ie  square  of  the  niimher  which  measures  the 
hypote^iuse  of  a  right  triangle  is  equal  to  the  sum  of  the 
squares  of  the  numbers  which  measure  the  other  two  sides. 

Hy]).  Let  ABC  be  a  rt.  A ,  with  rt. 
/  at  A. 

To  prove  BC'  =  AB'  +  AO'. 

Proof     Draw  the  ±  AD. 

Then  AB'  =  BC  X  BD,  ^ 

and 


(324) 
(324) 

Q.E.D. 


AC  =:  BC^  CD.  : 

Adding,  AB'  +  AC'  =  BC(BD  +  CD)  =  BC". 

328.  ScH.  1.  By  this  theorem  one  of  the  sides  of  a  right 
triangle  can  be  found  when  the  other  sides  are  known.  If 
we  know  the  two  sides  b  and  c  about  the  rt.  Z ,  the  hy- 
potenuse a  is  given  by  the  formula 

a''  =  b''  +  c\     .'.  a  =  Vb'  +  c\ 
Thus,  if  ^>  =  3,  c  =  4,  we  have  a  =  V9~+16  =  V2b  =  5. 
If  the  hypotenuse  a  and  one  of  the  sides  b  of  the  rt.  / 
be  known,  the  other  side  c  is  found  by  the  formula 


c'  =  a''-  b' 


Va'  -  W 


Thus,  for  rt  =  5,  and  Z>  =  3,  we  have  c  —  V25  —  9 
=  Vl6  =  4. 

329.  ScH.  2.  If  AC  is  the  diagonal  of  a 

square  ABCD,  we  have  (327) 

AC'  =  AB'  +  BC',  or  AC'  =  2  AB'. 

AC"      _        ,  AC       ^^ 

,\  -=ri  —  2,  and  -t-r  =  ^^• 
AB  AB 

Thus,  the  diago7ial  and  side  of  a  square  are  two  incom- 
mensurable lines,  since  their  ratio,  the  square  root  of  2,  is 
an  incommensurable  number. 


BOOK  III.— NUMERICAL  MEASUllES  IN  A  TRIANOLE.  161 

Note.— To  simplify  the  enunciations  of  the  following  theorems,  the  term 
square  of  a  line  will  be  given  to  the  second  power  of  the  number  which  ex- 
presses the  measure  of  the  line.  The  term  product  of  two  lines  will  be  given 
to  the  product  of  two  numbers  which  express  the  magnitude  of  those  lines, 
measured  with  the  same  unit. 


Proposition  26.    Theorem. 

330.  In  any  triangle,  the  square  of  the  side  opposite 
an  acute  angle  is  equal  to  the  sum  of  the  squares  of 
the  other  two  sides  di7ninished  hg  tivice  the  product  of 
one  of  these  sides  by  the  projection  of  the  other  side 
upon  it. 


Hyp.  Let  B  be  an  acute  Z  of  the  A 
ABC,  and  BD  the  projection  of  AB 
upon  BC. 

To  prove 

AC'  =  AB'  +  BC'  -  2BCx  BD. 
Proof.  Because  ADC  is  a  rt.  A, 
.-.  AC' =  AD' +  DC',  (327) 

=  AD'  +  (BC-BDr,(fig.l) 


or 


=  AD''  +  (BD  -  BC)''  (fig.  2) 


BCf  —  2  BC  X  BD,  by  Algebra*, 


BC  -  2BC  X  BD. 


(327) 

Q.E.D. 


*  The  square  of  the  difference  of  two  numbers  is  equal  to  the  sum  of  the 
squares  of  the  two  numbers  diminished  by  twice  their  product. 


162  PLANE  GEOMETRY. 

Proposition  27.    Theorem. 

331.  In  an  obtuse-angled  triangle,  the  square  of  the 
side  opposite  the  obtuse  angle  is  equal  to  the  sum  of  the 
squares  of  the  other  tiuo  sides  increased  by  twice  the  prod- 
uct of  one  of  these  sides  by  the  projection  of  the  other  side 
upon  it. 


C  B  D 

Hyp.  Let  B  be  the  obtuse  Z  of  the  A  ABO,  and  BD  the 
projection  of  AB  upon  CB  produced. 

To  prove  !(?  =  AB"  +  BO'  +  2BC  X  BD. 
Proof        CD  =  OB  +  BD. 

.  • .  CD'  =  CB'  +  BD'  +  2B0  X  BD.     (Algebra*) 
Add  AD  to  both  sides. 

AC'  =r  AB'  +  So'  +  2BC  X  BD.  (327) 

Q.E.D. 

332.  Cor.  From  the  three  p>receding  theorems,  it  follows 
that  the  square  of  the  side  of  a  triangle  is  less  than,  equal 
to,  or  greater  than,  the  sum  of  the  squares  of  the  other  two 
sides,  according  as  the  angle  opposite  this  side  is  acute, 
right,  or  obtuse. 

Note.— By  means  of  these  three  theorems  we  may  compute  the  altitudes  of  a 
triangle  whea  the  three  sides  are  known. 

EXERCISES. 

^     1.  If  the  side  BC  of  an  equilateral  triangle  ABC  be  pro- 
/duced  to  D,  prove  that  the  square  on  AD  is  equal  to  the 

squares  on  DC  and  CA  with  the  product  of  DC  and  CA. 
2.  If  ABC  be  an  isosceles  triangle  having  AB  =  AC, 

and  if  D  be  taken  in  AC  so  that  BD  =  BC,  prove  that  the 

square  on  BC  =  AC  X  CD. 

*  The  square  of  the  sum  of  two  numbers  is  equal  to  the  sum  of  the  squares 
of  the  two  numbers  increased  by  twice  their  product, 


BOOK  IIL-NUMEIUGAL  MEASURES  IN  A  TRIANGLE.  163 


I 


Proposition  28.    Theorem. 

333.  The  sum  of  tlie  squa7'es  on  two  sides  of  a  tricmgle 
is  equal  to  twice  the  square  on  half  the  third  side  increased 
hy  tioice  the  square  on  the  median  upon  that  side. 


} 


Hyp,  Let  ABC  be  a  A,  AE  the  median  bisecting  BC. 
To  prove  AB'  +  AC'  =  2  BE'  +  2AE'. 
Proof  Draw  AD  _L  to  BO. 

Then  one  of  the  Z  s  AEB,  AEG  must  be  obtuse  and  the 
other  acute :  let  Z  AEO  be  acute. 

In  the  aAEB,Xb'=BE'+AE'+  2BExED.(l)  (331) 
In.the  A AEC,  AC'=  CE'+ AE'-  20ExED.  (2)  (330) 
Adding,  and  observing  that  BE  =  CE,  we  get 

AB'+ AC'=  2  BE'+  2  AE'.  Q.E.D. 

334.  CoK.  Subtracting  (2)  from  (1)  in  (333),  we  have 

AB'-A0'=2BCxED. 

Hence,  the  difference  of  the  squares  on  two  sides  of  a  tri- 
angle is  equal  to  twice  the  product  of  the  third  side  by  the 
projection  ofihe  median  upon  that  side. 

Let  the  student  prove  the  case  in  which  AD  falls  without 
the  triangle  ABC,  or  coincides  with  AE. 

Note.— By  means  of  this  theorem  we  may  compute  the  medians  of  a  triangle 
when  the  three  sides  are  known. 


164  PLANE  GEOMETRY. 

Proposition  29.    Theorem. 

335.  If  two  cliords  cut  each  other  m  a  circle,  the  product 
of  the  segments  of  the  one  is  equal  to  the  product  of  the  seg- 
ments of  the  other, 

D 


Hyp,  Let  the  chords  AB,  CD  cut  at  P. 
To  prove  AP  X  BP  =  CP  X  DP. 

Proof  Join  AD  and  BC. 
In  the  As  APD,  CPB, 

Z  ADC  =  Z  ABC,     and     Z  BAD  =  Z  BCD, 
As  in  the  same  segment  of  a  ©  are  equal  (239). 

.  • .  A  APD  is  similar  to  A  CPB.  (309) 

.-.  AP:CP  =  DP:BP.  (312) 

.-.  APx  BP  =  CPX  DP.  (281) 

Q.E.D. 

336.  Def.  When  four  quantities,  such  as  the  sides  about 
two  angles,  are  so  related  that  a  side  of  the  first  is  to  a  side 
of  the  second  as  tlie  remaining  side  of  the  second  is  to  the 
remaining  side  of  the  first,  the  sides  are  said  to  be  recipro- 
cally proportional.     Therefore — 

337.  Cor.  1.  If  tioo  chords  cut  each  other  in  a  circle, 
their  segments  are  reciprocally  proportional. 

338.  Cor.  2.  //'  through  a  fixed  point  within  a  circle  any 
number  of  chords  are  draion,  the  products  of  their  segments 
are  all  equal, 

EXERCISES. 

1.  If  AP  =  4,  BP  =  5,  and  CD  =  12,  find  the  lengths  of 
CP  and  DP. 

2.  If  AB  =  20,  and  CD  =  24,  find  tlie  lengths  of  CP  and 
DP  when  CD  bisects  AB. 


BOOK  IIL— NUMERICAL  MEASURES  IN  A  TRIANGLE.  165 


Proposition  30.    Theorem. 

339.  If  from  a  point  without  a  circle  a  tangent  and  a 
secant  be  drawn,  the  tangent  is  a  mean  proportional  be- 
tween the  whole  secant  and  the  external  segment. 

Hyp.  Let  PC  and  PB  be  a  tangent  and 
a  secant  drawn  from  the  point  P  to  the 
OABO.  ^- 

To  prove  PB  :  PC  =  PC  :  PA. 

Proof     Join  CA  and  CB. 

IntheAsPAC,  PBC, 

ZACP=  ZABC, 

eacJi  being  measured  by  i  arc  A  C  (238),  (243). 

Z  P  is  common. 
.-.AS  PAC  and  PBC  are  similar.  (309) 

.-.PB:  PC  ==  PC:  PA.  (312) 

Q.E.D. 

340.  Cor.  1.  PC"  =  PB  x  PA. 

Therefore,  the  square  of  the  tangent  is  equal  to  the  prod- 
uct of  the  2vhole  secant  and  the  external  segment, 

341.  Cor.  2.  Since  PB  x  PA  =  PC',  (340) 
and                             PB  X  PD  =  PC',  (340) 

.-.PBx  PA^PExPD.  (Ax.  1) 

a 


Therefore,  if  from  a  point  without  a  circle  two  secants  he 
drawn,  the  product  of  one  secant  and  its  external  seg^nent 
is  equal  to  the  product  of  the  other  and  its  external  segment. 

342.  Cor.  3.  If  from  a  point  without  a  circle  any  num- 
ber of  secants  are  drawn,  the  products  of  the  lohole  secants 
and  their  external  segments  are  all  equal. 


.^.-.•* 


166  PLANE  GEOMETRY. 

Proposition  3 1 .    Theorem. 

343.  If  any  angle  of  a  triangle  is  bisected  hy  a  straight 
line  wliicli  cuts  the  base,  the  product  of  the  tioo  sides  is  equal 
to  the  product  of  the  segments  of  the  base  plus  the  square  of 
the  bisector. 

Hyp.  In  the  A  ABC  let  AD  bisect  the 
Z  BAG. 

To  prove  AB  X  AC  =  BD  x  DC  +  AD\ 

Proof  Describe  a  O  about  the  A  ABC, 
and  produce  AD  to  meet  tlie  Oce  in  E. 
Join  CE.  E 

Then  in  the  As  BAD,  EAC, 

ZBAD=ZEAC.  (Hyp.) 

ZABD  =  ZAEC. 

Zs  in  ilie  same  segment  of  a  Q  are  equal  (239). 

.*.  AS  BAD  and  EAC  are  similar.  (309) 

.-.  AB  :  AE  =  AD  :  AC.  (312) 

.-.  AB  X  AC  =:  AE  X  AD  (281) 

=  (ED  +  AD)  AD 

'   =  ED  X  AD  +  AD'. 

But       ED  X  AD  =  BD  X  DC.  (335) 

.-.  AB  X  AC  =  BD  X  DC  +  AD'.  q.e.d. 

Note.— By  means  of  this  theorem  we  may  compute  the  lengths  of  the  bisect- 
ors of  the  angles  of  a  triangle  when  the  three  sides  are  known. 

EXERCISE. 

If  the  vertical  angle  BAC  be  externally  bisected  by  a 
straight  line  which  meets  the  base  in  D,  show  that  the  prod- 
uct of  AB,  AC  together  with  the  square  on  AD  is  equal  to 
the  product  of  the  segments  of  the  base. 


BOOKlIL—NUMEUWALMEASUUESmA  TIUAMULE.  167 


Proposition  32.    Tiieorem. 

344.  The  "product  of  two  sides  of  a  triangle  is  equal  to 
\e  product  of  the  diameter  of  the  circumscribed  circle  by 
the  perpendicular  let  fall  upon  the  third  side  from  the  vertex 
of  the  opposite  angle. 

Hyp,  In  the  A  ABC  lefc  AD  be  J.  to 
BO,  and  AE  the  diameter  of  the  circum- 
scribed O. 

To  prove    AB  x  AC  =  AE  x  AD. 

Proof.  In  the  AS  ABD,  AEC, 

ZB=ZE, 

each  being  measured  by  i  arc  AC  (238), 

and       Z  ADB  =  Z  ACE  =  a  rt.  Z .     (Cons.)  and  (240) 

.-.    AS  ABD,  AEC  are  similar.  (309) 

.-.  AB:  AE  =  AD:  AC.  (312) 

.-.  AB  X  AC  =  AE  X  AD.  (281) 

Q.E.I). 

Note.— By  means  of  this  theorem  we  may  compute  the  radius  of  the  circle 
circumscribed  about  a  triangle  when  the  three  sides  are  known. 

EXERCISE. 

The  product  of  the  two  diagonals  of  a  quadrilateral  in- 
scribed in  a  circle  is  equal  to  the  sum  of  the  products  of  its 
opposite  sides. 

SuaoKSTioN.— Make  /DAE  =  /BAG. 
Then  in  the  similar  a  s  DAE,  CAB, 

AD  ;  AC  =  DE  :  BC;    .-.  AD  .  BC  =  AC  DE. 
Also  in  similar  as  DAC,  EAB, 

AB:AC  =  BE:CD;    .-.  AB  .  CD  =  AC.  BE. 
Add.  AD  .  BC  +  AB  .  CD  =  AC  .  DB. 


168 


PLANE  GEOMETRY. 


Problems  of  Constructiq]^. 

Proposition  33.    Problem. 

345.  To  divide  a  given  straigld  line  into  parts  projjor- 
tio7ial  to  given  straigld  lines. 

Given,  the  line  AB,  and   the   lines   P   N  M 
M,  N,  P. 

Reqiiiredf  to  divide  AB  into  parts 
proportional  to  M,  N,  P. 

Cons.  From  A  draw  the  indefinite 
straight  line  AH  making  any  /  with 
AB. 

On  AH  take 


DE  =  P. 


AC  =  M,     CD  =  N, 

Join  EB,  and   through   D,  C   draw 
DG,  CF  II  to  EB. 

Then  AB  is  divided  at  G  and  F  into  parts  proportional 
to  M,  N,  P. 

Proof,  Since  CF,  DG,  EB  are  || ,  (Cons.) 

AF  _FG_GB 
•'•  AC"~CD~DE* 
If  two  St.  lines  are  cut  by  any  number  of  \\8,  the  corresponding  segments 
are  proportional  (300). 

ButAC  =  M,     CD  =  ]Sr,     DE  =  P.  (Cons.) 

AF       FG      GB 

346.  ScH.  If  it  be  required  to  divide  a  line  AB  into  any 
number  of  equal  parts,  take  the  same  number  of  equal  parts 
on  AH,  so  that  AC  =  CD  =  DE,  and  complete  the  con- 
struction as  before ;  then  AF  =  FG  =  GB. 

EXERCISES. 

1.  Divide  a  straight  line  into  five  equal  parts. 

2.  Give  the  construction  for  cutting  ofl:  two-sevenths  of 
a  given  straight  line. 


BOOK  IIL— PROBLEMS  OF  CONSTRUCTION.        169 


Proposition  34.    Problem. 

347.   To  find  a  fourth  proportional  to  three  given  straight 
lines. 


Given,  the  three  lines  M,  N,  P. 

Required,  to  find  a  fourth  propor- 
tional to  M,  N,  P. 

Cons.  Draw  the  two  indefinite  straight 
lines  AG,  AH,  making  any  /  with  each 
other. 

On  AH  take  AB  =  M,  BC  =  N  ;  and 
on  AG  take        AD  =  P. 

Join  DB,  and  through  0  draw  CE  II  to 
DB. 

Then  DE  is  the  fourth  proportional 
required. 

Proof.  Since  BD  and  CE  are  II , 

.-.  AB:BO  =  AD:DE. 

A  St.  L  \\to  a  side  of  a  A  divides  the  other  two  sides  proportionally  (298). 

But  AB  =  M,  BC  =  ^,  AD  =  P.  (Cons.) 

.-.  M:ISr  =  P:DE.  q.e.f. 

348.  ScH.  If  the  lines  IST  and  P  are  equal,  then  BC  and 
AD  are  both  laid  off  equal  to  N,  and  DE  is  the  third  pro- 
portional to  M  and  N  (280).  The  proportion  in  (347)  then 
becomes 

M  :  N  =  N  :  DE. 

EXERCISES. 

1.  Construct  a  fourth  proportional  to  the  lines  3,  7,  11. 

2.  If  from  D,  one  of  the  angles  of  a  parallelogram  ABOD, 
a  straight  line  is  drawn  meeting  AB  at  E  and  CB  produced 
at  F;  show  that  OF  is  a  fourth  proportional  to  EA,  AD, 
mad  AB. 


170  PLANE  GEOMETRT. 


Proposition  35.    Problem. 

349.  To  find  a  mean  proportional  between  tiuo  given 
straight  lines. 

Given,  the  lines  M  and  N.  M 

Reqicired,  to   find  a  mean  propor-     "^ 
tional  between  M  and  N. 

Cons.  On  an  indefinite  straight  line 
take  AD  =  M,  and  DB  =  N. 


I  \ 


On  AB  describe  a  semicircle,  and  i  "Twr — ~U~^\" 
draw  DO  J.  to  AB. 

Then  DC  is  the  mean  proportional  required. 

Proof,  Since         AD:DC  =  DC:DB,  (325) 

and                            AD  =  M,  DB  =  N,          >  (Cons.) 

.-.  M:DC  =  DC:]Sr.  q.e.f. 

350.  ScH.  The  mean  proportional  between  two  lines  is 
often  called  the  geometric  mean,  while  their  half  sum  is 
called  the  arithmetic  mean. 

EXERCISE. 

From  a  given  point  P  in  a  circle  a  perpendicular  PM  is 
drawn  to  a  given  chord  AB ;  from  A,  B  perpendiculars  AC, 
BD  are  drawn  to  the  tangent  at  P:  prove  that  PM  is  a  mean 
proportional  between  AC  and  BD. 

351.  Def.  a  straight  line  is  said  to  be  divided  in 
extreme  and  mean  ratio,  when  the  whole  is  to  the  greater 
segment  as  the  greater  segment  is  to  the  less. 

Thus,  the  line  AB  is  divided  in  extreme 

and  mean  ratio  at  C  if  <^  C B 

AB  :  AC  =  AC  :  CB. 


BOOK  IIL-PROBLEMS  OF  CONSTRUCTION.        171 

Proposition  36.    Problem. 

352.  To  divide  a  given  straight  line  in  extreme  and 
7nean  ratio. 

Given,  the  st.  line  AB.  /'"*      *"*'v. 

Required,  to  divide  it  in  extreme  /            ,,''''^f  | 

and  mean  ratio.  i      ,.     ' 

Cons.  At  B  erect  the  J.  BO =i  AB.  /i^<' 

With  centre  C  and  radius  CB  de-     .--'''^ 
V.        ^  A  F  B 

scribe  a  O. 

Join  AC,  cutting  the  O  at  J),  and  produce  it  to  meet 
the  O  again  at  E. 

On  AB  lay  off  AF  =  AD. 

Then  F  is  the  pt.  of  division  required. 
Proof.   Since  AB  is  a  tangent  and  AE  a  secant  to  the  O, 
.  • .  AE  :  AB  =  AB  :  AD.  (1) 

The  tangent  is  a  mean  pt'opoo^tional  between  the  wJiole  secant  and  the 

external  segment  (339). 

.-.  by  division,  AE  -  AB  :  AB  =  AB  -  AD  :  AD.         (288) 

But  AB  =  DE  and  AD  =  AF.  (Cons.) 

.-.  AE  -  AB  =  AD  =  AF,  and  AB  -  AD  =  FB. 

Substituting  these  values  in  the  above  proportion,  we  have 

AF  :  AB  =  FB  :  AF.  4^^ 

.-.  AB:AF  =  AF:FB.  ^-y^s) 

. ' .  AB  is  divided  in  extreme  and  mean  ratio  at  F.       q.e.f. 

353.  ScH.  If  BA  be  produced  to  the  left  of  A  to  a 
point  F'  so  that  AF'  =  AE,  then  F'  is  another  point  of 
division  having  the  same  property  as  F. 

Thus  we  have  from  proportion  (1)  in  (352),  by  composition, 

AE  H-  AB  :  AE  =  AB  +  AD  :  AB.  (287) 

But        AE  +  AB  =  BF',  and  AB  -[-  AD  =  AF'.  (Cons.) 

.-.  BF':  AF'  =  AF':  AB, 
or  AB  :  AF'  =  AF' :  F'B. 

.  • .  AB  is  divided  in  extreme  and  mean  ratio  at  F'. 
AB  is  said  to  be  divided  at  F  internally,  and  at  F'  ex- 
terjially,  in  extreme  and  mean  ratio. 

Note.— This  division  of  the  straight  line  was  called  by  the  ancient  geometers 
the  golden  section. 


172  PLANE  GEOMETRY. 


Proposition  37.    Problem. 

354.   On  a  given  straiglit  line  to  construct  a  2)olygon 
similar  to  a  given  ^polygon. 

Given,  the  polygon  ABODE  and 
the  straight  line  A'B'. 

Required,  to  construct  on  A'B'  a 
polygon  similar  to  ABODE. 

Cons.  Let  AB  be  the  side  of  the 
given  polygon  to  which  A'B'  is  to 
be   homologous  ;  join  AO  and  AD. 

At  A'  make  Z  B'A'O'  =  /  BAO, 
and  at  B'  make  Z  A'B'O'  =  Z  ABO. 

Also  at  A'  makez  O'A'D'  =  Z  CAD, 
and  at  0'  make  Z  A'O'D'  =  z  AOD. 

Also  at  A'  make  Z  D'A'E'  =  Z  DAE, 
and  at  D'  make  Z  A'D'E'  =  z  ADE. 

Then  A'B'O'D'E'  is  the  required  polygon. 
Proof.  Since  the  A  s  ABO,  A'B'O'  are  mutually  equi- 
angular, (Ooiis.) 
.*.  they  are  similar.  (309) 
Also  AS  AOD,  A'O'D'  are  similar.  (309) 
Likewise  A  s  ADE,  A'D'E'  are  similar,                     (309) 

.  •.  the  polygons  ABODE,  A'B'O'D'E'  are  similar, 
being  composed  of  ttie  same  number  of  as  similar  each  to  each,  ami 

similarly  placed  ( 3^0) .  q. e. f. 


BOOK  IlI.-APPLIGATIONS.  173 

Applications. 

1.  To  find  the  altitudes  of  a  triangle  in  terms  of  its  sides.* 
See  (332). 

Let  ABC  be  a  A  ;  denote  by  a,  l,  c, 
the  lengths  of  the  sides  opposite  the  ^/^ 

Z  s  A,  B,  C,  respectively,  and  by  h  the  \cy^    I 

altitude  AD.  y^  A 

Of  the  two  Z  s  B  and  C,  at  least  one  is     Z^ / 

acute  ;  suppose  it  to  be  the  Z  C.    Then : 


In  the 

aADC,  A^ 

=  h'  -  CD'. 

(328) 

In  the 

A  ABC,  6'^ 

^a'  j^¥  -  2a  X  CD. 

(330) 

.-.CD 

2a 

r.V  = 

{a^-\-l^-oy 

^.a'h'-ia'+h'-cy 

■^              4^' 

~                4a" 

{2ah^-a^W-c'){2al-ce-V^-c^)               ,,^  ,, 

4.a'                                   ^^ 

J   -^^^S'  1  / 

-{a-hy\ 

.  {a+i-^c){a+h- 

c)(c-\-a-'b)(c-a-\-y) 

(1) 

4a' 

Put 

a+b  + 

c  =  2s. 

Then 

a-\-b- 

c  =  2{s  —  c). 

c  -\-  a  — 

■b  =  2{s-  I). 

c  —  a  -\- 

h  =  2(s  -  a). 

Substituting  in  (1)  and 

extracting  the  square 

root,  we 

,        2 

/'  =  -   Ms  -  a){s  -  b){s  -  c). 

2.  To  find  the  medians  of  a  triangle  in  terms  of  its  sides. 
See  (333). 

*  See  Trait6  de  G6otn6trie,  par  Roiich6  e4;  Comberousse,  p.  139. 
t  The  difference  of  the  squares  of  two  numbers  equals  the  product  of  the  pum 
«ind  difference  of  the  two  numbers, 


174  PLANE  GEOMETRY. 

Denote  by  m  the  median  on  the  side  BO. 

Then,  ■!)'-{- c' =^ 'Zm' ^  ^(^^\  (333) 

r,m  =  ^  i/2(^'  +  c')  -  a\ 
3.  To  find  the  bisectors  of  a  triangle  in  terms  of  its  sides. 
See  (343.) 

Denote  the  bisector  AD,  figure  of  (343),  by  d,  and  the 
sides  as  in  Exs.  1  and  2. 

By  (343),     ^c  =  BD  X  DC  +  d\ 

.•.^•^  =  Z»c~BD  X  DC.  (1) 

By  (303),  BD  ^  DC  ^  BD^C  ^     a     _ 

•^^^c  b  b  -\-c  b  -\-c  ^       ' 

...BD  =  ^^,andDC:=     ^* 


b  +  o'  b-\-c 

Substituting  these  values  in  (1)  and  reducing,  and  de- 
noting by  s  the  semi-perimeter  of  the  A ,  we  have 

4.  To  find  the  radius  of  the  circumscribed  circle  in  terms 
of  the  sides  of  the  triangle.     See  (344). 

Denote  the  radius  by  R,  figure  of  (344),  and  the  sides  as 
before. 

By  (344),    bc  =  2^x  AD. 

Substituting  for  AD  its  value  in  Ex.  1,  and  solving  for  E, 
we  have 

P  _  abc 

~  4  Vs{s-a){s-l)){s  -  v) ' 

5.  Find  the  altitudes  of  a  triangle  whose  sides  are  13,  9, 
and  6.  Ans.  3.G41,  5.259,  7.88G. 

6.  Find  the  medians  of  the  above  triangle. 

Ans.  4.031,  9.069,  10.770. 

7.  Find  (1)  the  bisectors  of  the  angles  of  the  above  tri- 
angle, and  (2)  the  radius  of  the  circumscribed  circle. 

Ans,   (1)  3.833,  7.778,  I0.407j  (2)  7.416. 


BOOK  IIL-EXERCI8ES.     THEOREMS.  175 

exercises. 
Theokems. 

1.  If  two  circles  touch  each  other,  either  internally  or 
externally,  any  two  straight  lines  drawn  from  the  point  of 
contact  will  be  cut  proportionally  by  the  circumferences. 

2.  If  two  circles  intersect  in  P,  and  the  tangents  at  P  to 
the  two  circles  meet  the  circles  again  in  Q  and  E;  prove 
that  PQ :  PR  in  the  same  ratio  as  the  radii  of  the  circles. 

3.  ABO,  DEF  are  two  isosceles  triangles,  BO,  EF  being 
the  bases.  If  AB  :  B0^=  DE  :  EF,  show  that  the  triangles 
are  similar. 

4.  P  is  a  point  in  a  diagonal  of  a  parallelogram.  EPF, 
GPH  join  points  in  the  opposite  sides  of  the  parallelogram. 
EH,  GF  cut  the  diagonal  in  which  P  is.  Show  that  EH  is 
parallel  to  GF. 

5.  The  parallel  straight  lines  AB,  OD  are  joined  by  AD, 
BO  which  intersect  in  0 ;  in  AB,  OD  points  E,  F  are  taken 
such  that  AE  :  EB  =  DF  :  FO  :  prove  that  E,  0,  F  are  col- 
linear.     (1G4). 

C.  The  side  BO  of  a  triangle  ABO  is  bisected  at  D,  and 
the  angles  ADB,  ADO  are  bisected  by  the  straight  lines 
DE,  DF,  meeting  AB,  AO  at  E,  F  respectively :  show  that 
EF  is  parallel  to  BO. 

Apply  (303). 

7.  AB  is  a  diameter  of  a  circle,  OD  is  a  chord  at  right 
angles  to  it,  and  E  is  any  point  in  OD;  AE  and  BE  are 
drawn  and  produced  to  cut  the  circle  at  F  and  G;  show 
that  the  quadrilateral  OFDG  has  any  two  of  its  sides  in  the 
same  ratio  as.  the  remaining  two. 

Apply  (303). 

8.  In  the  circumference  of  the  circle  of  which  AB  is  the 
diameter,  take  any  point  P,  and  draw  PO,  PI)  on  opposite 
sides  of  AP,  and  equally  inclined  to  it,  meeting  AB  at  0 
and  D:  show  that  AO  :  BO  =  AD  :  BD. 

Apply  (303)  and  (304). 

9.  From  tlie  same  point  A  straight  lines  are  drawn  mak- 
ing the  angles  BAO,  OAD,  DAE  each  equal  to  half  a* right 
angle,  and  they  are  cut  by  a  straight  line  BODE,  which 


176  PLANE  GEOMETliY. 

makes  BAE  an  isosceles  triangle :  show  that  BC  or  DE  is  a 
mean  proportional  between  BE  and  CD. 

Apply  (303)  and  (304). 

10.  ABCD  is  a  quadrilateral  having  two  of  its  sides  AB, 
CD  parallel;  AF,  CG  are  drawn  parallel  to  each  other 
meeting  BC,  AD  respectively  in  F,  G.  Prove  that  BG  is 
parallel  to  DF. 

11.  If  D  be  the  middle  point  of  the  side  BC  of  a  trian- 
gle ABC,  and  if  any  straight  line  be  drawn  through  C  meet- 
ing AD  in  E  and  AB  in  F;  show  that  the  ratio  of  AE  to 
ED  will  be  twice  the  ratio  of  AF  to  FB. 

12.  If  0  be  the  centre  of  the  inscribed  circle  of  the  tri- 
angle ABC,  and  AO  meet  BC  in  D;  prove  that  AO  is  to 
OD  as  the  sum  of  AB  and  AC  is  to  BC. 

Apply  (303). 

13.  ABCD  is  a  quadrilateral  inscribed  in  a  circle;  AD, 
BC  meet  in  P,  and  the  angle  P  is  bisected  by  a  straight 
line  cutting  AB,  CD  in  E,  F.  Show  that  AB  :  BE  = 
CD  :  DF. 

Apply  (309). 

14.  ABC  is  a  straight  line,  and  ABD,  BCE  triangles  on 
the  same  side  of  it,  such  that  the  angles  ABD,  EBC  are 
equal,  and  AB:  BC  =  BE:  BD:  if  AE,  CD  intersect  in  F, 
prove  that  AFC  is  an  isosceles  triangle. 

See  (314). 

15.  Two  circles  touch  in  C,  a  point  D  is  taken  outside 
them  such  that  the  radii  AC,  BC  subtend  equal  angles  at 
D,  and  DE,  DF  are  tangents  to  the  circles :  if  EF  cut  DG 
in  G,  prove  that  DE  :  DF  =  EG  :  GF. 

16.  C  is  the  centre  of  a  circle,  and  A  any  point  within 
it  ;  CA  is  produced  through  A  to  a  point  B  such  that  the 
radius  is  a  mean  proportional  between  CA  and  CB  :  show 
that  if  P  be  any  point  on  the  circumference,  the  angles 
CPA  and  CPB  are  equal. 

See  (314). 

17.  AB,  AC  are  the  equal  sides  of  an  isosceles  triangle ; 
the  straight  line  bisecting  AB  at  right  angles  meets  BC  in 
D  :  prove  that  the  square  on  AB  =  BC  X  BD. 


BOOK  III. -EXERCISES.     THEOREMS.  177 

18.  The  square  on  the  base  of  an  isosceles  triangle  is  equal 
to  twice  the  product  of  either  side  by  the  part  of  that  side 
intercepted  between  the  perpendicular  let  fall  on  the  side 
from  the  opposite  angle  and  the  end  of  the  base. 

Apply  (330.) 

19.  ABC  is  a  triangle  having  the  sides  AB,  AC  equal;  if 
AB  is  produced  to  D  so  that  BD  is  equal  to  AB,  show  that 
the  square  on  CD  is  equal  to  the  square  on  AB",  together 
with  twice  the  square  on  BC. 

20.  The  sum  of  the  squares  on  the  four  sides  of  a  paral- 
lelogram is  equal  to  the  sum  of  the  squares  on  the  diagonals. 

Apply  (333). 

21.  The  base  of  a  triangle  is  given  and  is  bisected  by  the 
centre  of  a  given  circle  :  if  the  vertex  be  at  any  point  of  the 
circumference,  show  that  the  sum  of  the  squares  on  the  two 
sides  of  the  triangle  is  constant. 

22.  In  any  equilateral  the  sum  of  the  squares  on  the  diag- 
onals is  equal  to  the  sum  of  the  squares  on  the  straight  lines 
joining  the  middle  points  of  opposite  sides. 

Apply  (155)  and  then  Ex.  20. 

23.  The  sum  of  the  squares  on  the  four  sides  of  a  quad- 
rilateral is  equal  to  the  sum  of  the  squares  on  the  diagonals, 
increased  by  four  times  the  square  on  the  line  joining  the 
middle  points  of  the  diagonals. 

Apply  (333). 

24.  AB  is  a  diameter  of  a  circle,  the  chords  AC,  BD 
intersect  in  a  point  E  inside  the  circle  :  prove  that  the 
square  on  AB  is  equal  to  the  sum  of  the  products  of  AC, 
AE  and  BD,  BE. 

Apply  (335)  and  (331). 

25.  AD,  BE,  CF,  the  perpendiculars  from  the  vertices 
on  the  opposite  sides  of  a  triangle  ABC,  intersect  in  0  : 
prove  that  the  products  of  AO,  OD  and  BO,  OE  and  CO, 
OF  are  equal  to  each  otlier. 

Apply  (335). 

26.  With  a  point  0  inside  a  circle  ADC  as  centre  a  circle 
is  described  cutting  the  former  circle  in  D,  E  ;  a  chord 
AOB  is  drawn  to  the  first  circle  :  if  the  product  AO,  OB  is 


178  •  PLANE  GEOMETRY, 

equal  to  the  square  on  the  radius  of  the  second  circle,  prove 
that  D,  0,  E  are  collinear. 

Apply  (335). 

27.  If  two  circles  intersect,  prove  that  tangents  drawn  to 
them  from  any  point  in  their  common  chord  produced  are 
equal  to  each  other. 

28.  AEB,  PEDQ  are  respectively  a  diameter  and  a  chord 
of  a  circle  at  right  angles  to  each  other,  ADC,  BC  are  otlier 
chords.  Prove  that  the  product  of  AB,  AE  is  equal  to  the 
product  of  AC,  AD.     (341). 

29.  If  AB  be  a  diameter  of  a  circle,  and  APQ  a  straight 
line  cutting  the  circle  again  at  P  and  a  fixed  straight  line 
perpendicular  to  AB  at  Q;  the  product  of  AP,  AQ  is  con- 
stant.    (341). 

30.  A,  B,  D  are  three  points  in  order  on  a  circle ;  C  is 
the  middle  point  of  the  arc  AB ;  if  AC,  AD,  BD,  CD  be 
joined,  prove  that  the  square  on  CD  is  equal  to  the  square 
on  AC  with  the  product  of  AD,  BD. 

31.  A  circle  is  described  round  an  equilateral  triangle, 
and  from  any  point  in  the  circumference  straight  lines  are 
drawn  to  the  angular  points  of  the  triangle :  show  that  one 
of  these  straight  lines  is  equal  to  the  sum  of  the  other  two. 

32.  From  the  extremities  B,  C  of  the  base  of  an  isosceles 
triangle  ABC,  straight  lines  are  drawn  at  right  angles  to 
AB,  AC  respectively,  and  intersecting  at  D :  show  that 
BC  X  AD  =  2  AD  X  DB. 

Problems. 

33.  AD  bisects  the  angle  BAC.  From  any  point  in  AB 
draw  a  straight  line  to  AC  which  shall  be  divided  by  AD 
in  a  given  ratio,  i.e.,  into  parts  which  have  the  same  ratio 
which  two  given  straight  lines  have  to  each  other. 

34.  Through  a  given  point  between  two  given  straight 
lines  draw  a  straight  line  which  shall  be  terminated  by  the 
given  lines  and  divided  by  the  point  in  a  given  ratio. 

35.  Upon  a  given  portion  AC  of  the  diameter  AB  of  a 
semicircle  another  semicircle  is  described.     Draw  a  line 


BOOK  IIL—EXBHGISES.     PROBLEMS.  179 

throiigli  A  so  that  the  part  intercepted  between  the  semi- 
circles may  be  of  given  length. 

36.  ABC  is  a  segment  of  a  circle;  the  chord  AC  is 
divided  into  two  parts  at  D.  Divide  the  arc  ABC  into  two 
parts  so  that  the  chords  of  the  parts  shall  have  the  same 
ratio  which  AD,  DC  have  to  each  other. 

Complete  the  ®  ABCE  ;  bisect  arc  AC  opp.  B  in  E  ;  produce  ED  to  B,  etc. 

37.  Divide  the  straight  line  AB  into  two  parts  at  C  so 
that  the  square  on  AC  may  be  equal  to  twice  the  square  on  BC. 

Make  zBAD  =  J^  rt.  z,  draw  BD  ±  to  AB,  make  zADC  =  ZDAB,  etc. 

38.  From  a  given  point  as  centre  describe  a  circle  cutting 
a  given  straight  line  in  two  points,  so  that  the  product  of 
their  distances  from  a  fixed  point  in  the  straight  line  may 
be  equal  to  a  given  square. 

39.  ABC  is  a  given  obtuse-angled  triangle :  find  a  point 
D  in  the  side  BC  opposite  the  obtuse  angle,  so  that 
BD  X  DC  =  Ai5'. 

Let  O  be  the  cent,  of  the  ®  about  a  ABC.  Join  AO,  on  it  describe  |  ®  ADO, 
cutting  BC  in  D  ;  join  AD  and  produce  it  to  E,  etc. 

40.  Describe  a  circle  which  shall  pass  through  two  given 
points  and  touch  a  given  straight  line.  Two  solutions. 
(339). 

Let  A,  B  be  the  pts.,  XY  the  line  ;  produce  AB  to  meet  XY  in  C,  etc. 

41.  Describe  a  circle  to  pass  through  a  given  point  and 
touch  two  given  straight  lines  which  are  not  parallel  to  each 
other.     Two  solutions. 

Let  given  pt.  P  be  between  the  given  lines  AB,  AC  ;  bisect  Z  BAC,  etc. 

42.  From  a  given  point  on  the  circumference  of  a  given 
circle  draw  two  chords  so  as  to  be  in  a  given  ratio  and  to 
contain  a  given  angle. 

Let  P  be  the  given  pt.;  draw  diameter  PA  ;  make  Z  APB  =  given  z,  etc. 

43.  From  any  point  A  on  one  of  two  intersecting  straight 
lines  OA,  OB  draw  two  straight  lines  AB,  AC  cutting  OB 
ill  B  and  C,  such  that  the  triangles  AOB,  OAC  may  be 
similar,  and  the  sides  AB,  AC  have  a  given  ratio  to  each 
other. 

44.  To  describe  a  circle  which  shall  pass  through  two 
given  points  and  touch  a  given  circle. 


Book   IV. 
AEEAS'  OF  POLYGONS. 


Measurement  of  Areas. 

355.  Def.  The  area  of  a  surface  is  its  numerical  meas- 
ure (229),  referred  to  the  unit  of  surface. 

To  form  a  unit  of  surface,  we  take  any  unit  of  length, 
as  an  inch,  a  foot,  etc.,  and  construct  a  square  on  it. 
This  square  is  the  corresponding  unit  of  surface,  or  the 
superficial  unit. 

Two  surfaces  are  equivalent  when  they  have  equal  areas 
and  cannot  coincide;  they  are  equal  when  they  can  coin- 
cide. 

Proposition  1 .    Theorem. 

356.  If  two  rectangles  have  equal  altitudes,  their  areas 
are  to  each  other  as  their  bases. 

Hyp.  Let   ABCD,    AEHD   be  two  "^ 

rectangles,  having  equal  altitudes. 

ABCD  _  AB 
AEHD  ~  AK 


To  prove 


Case  I.  Wien  AB  and  AE  are  com- 
mensurahle. 

Proof.  Take  AF,  any  common  meas- 
ure of  AB  and  AE,  and  suppose  it  to 
be  contained  5  times  in  AB  and  4  times  in  AE. 

AB       5 
AE~4" 


j 

A     f 

I 

-                         B 
3                        H 

j       1 

1     i 

A      F 


Then 


Draw  J_s  to  AB  and  AE  at  the  several  points  of  division. 
The  rectangle  ABCD  will  be  divided  into  5  rectangles,  and 
AEHD  into  4  rectangles. 

180 


D 

KC 

1 

A 

QB 
D                         H 

BOOK  IV.-MEASUREMENT  OF  AREAS.  181 

These  rectangles  are  all  equal.  (136) 

ABCD  _  5 
•'•  AEHD~4' 

ABCD  _  AB 
•*•  AEHD~  AE*  (Ax.  1) 

Case  II.   When  AB  and  AB  are  incommensurable. 

Proof,  In  this  case  we  know  (232) 
that  we  may  always  find  a  line  AG 
as  nearlg  equal  as  ice  please  to  KB, 
and  such  that  AG  and  AE  are  com- 
mensurable. 

Draw  GK  II  to  BC. 

Then,  since  AG  and  AE  are  com- 
mensurable, 

AGKD      AG    ,n       Tx        A  E 

...^^jjj^  =  ^-^.(CaseI)        A 

Now,  these  two  ratios  being  always  equal  while  the  com- 
mon measure  is  indefinitely  diminished,  they  will  be  equal 
when  G  moves  up  to  and  as  nearly  as  lue  j^lcase  coincides 
with  B  (233). 

ABCD  _  AB 
•'•   AEHD~AE'  ^•^•''• 

357.  Cor.  Since  the  altitudes  of  rectangles  maybe  con- 
sidered as  their  bases,  and  their  bases  as  their  altitudes, 
therefore : 

The  areas  of  rectangles  having  equal  bases  are  to  each 
other  as  their  altitudes. 

Note —In  these  propositions  the  words  "rectangle,"  "triangle,"  etc.,  are 
often  used  to  denote  the  "area  of  the  rectangle,"  the  "area  of  the  triangle," 
etc. 

EXERCISES. 

The  area  of  a  rectangle  is  1728  square  feet,  and  its  base 
is  2  yards  :  what  is  the  area  of  a  second  rectangle  whose 
base  is  5  yards  and  whose  altitude  is  the  same  as  that  of 
the  first  ? 


182 


PLANE  GEOMETRY, 


Proposition  2.    Theorem. 

358.   The  areas  of  any  tivo  rectangles  are  to  each  other 
as  the  products  of  their  bases  by  their  altitudes. 

Hyp,  Let  R  and  R'  be  two 
rectangles,  b  and  Z>'  their  bases, 
a  and  a'  their  altitudes, 

^  U        a  X  b 

To  prove 


R' 


Xb'' 


c: 


1 


a  ^  ^ 

J  9 


Proof  Construct  the  rect- 
angle S  having  the  same  base 
b  as  that  of  R,  and  the  same 
altitude  a'  as  that  of  R'. 

Then  _^ 

D--      a' 

rectangles  of  equal  bases  are  to  each  other  as  tJwir  altitudes  (357) ; 

-S         b 
and  ^  =  y, 

rectangles  of  equal  altitudes  are  to  each  oilier  as  tlieir  bases  (356). 

Multiplying  these  ratios,  and  canceling  the  common  fac- 
tor, we  have 

R^_   a  X  b 

R'  ~ «'  X  y 

359.  ScH.  By  the  product  of  two  lines  is  meant  the 
product  of  the  numbers  which  represent  those  lines  when 
they  are  measured  by  a  common  linear  unit  (277). 

EXERCISE. 

Find  the  ratio  of  two  rectangles,  the  base  of  the  first 
bemg  3  yards  and  its  altitude  13  feet,  the  base  of  the  sec- 
ond being  8  feet  and  its  altitude  39  inches, 

Reduce  to  the  same  unit,  and  compare.  AnS,  4^. 


Q.E.D. 


BOOK  IV.- AREAS  OF  POLYGONS. 


183 


(358) 


Proposition  3.    TFieorem. 

360.  The  area  of  a  rectangle  is  equal  to  the  product  of 
its  base  and  altitude. 

Hyp,  Let  K  be  the  rectangle,  b  the 
base,  and  a  the  altitude  expressed  in 
numbers  of  the  same  linear  unit;  and 
let  S  be  the  square  whose  side  is  the 
linear  unit. 

To  prove  area  of  K  =  «  X  ^. 

Proof  ^-=__^^X^. 

But  since  S  is  the  unit  of  surface, 

.  • .  R  -f-  S  ==  the  area  of  R.  (355) 

.'.  area  of  R  =  a  X  b,  q.e.d. 

361.  ScH.  1.  The  statement  of  this  Proposition  is  an 
abbreviation  of  the  following: 

The  number  of  units  of  area  in  a  rectangular  fig^ire  is 
equal  to  the  product  of  the  number  of  linear  units  in  its 
base  by  the  number  of  linear  units  in  its  altitude. 

When  the  base  and  altitude  can  be    ^^  ^ 

expressed  exactly  in  terms  of  some  com- 
mon unit,  this  proposition  is  rendered 
evident  by  dividing  the  figure  into 
squares,  each  equal  to  the  unit  of  meas- 
ure. Thus,  if  AB  contain  4  linear 
units  and    AD  3,  and   if  through  the  ^  ^ 

points  of  division  parallels  are  drawn,  it  is  seen  that  the 
rectangle  is  divided  into  three  rows,  each  having  four 
squares,  or  into  four  coluvuis,  each  having  three  squares. 
Hence  the  whole  rectangle  contains  3  X  4,  or  12  squares, 
each  equal  to  the  unit  of  measure. 

Similarly,  if  AB  and  AD  contain  m  and  n  units  of  length 
respectively,  the  rectangle  will  contain  mil  units  of  area. 

362.  ScH.  2.  The  area  of  a  square  is  equal  to  the  second 
poumr  of  its  side,  being  the  product  of  two  equal  sides. 

In  Geometry,  the  expression  "  rectangle  of  two  lines  ■"  is  frequently  used  for 
the  "  product  of  two  lines,"  meaning  the  product  of  their  numerical  measures. 


! 


184  PLANE  OEOMETBY. 


Proposition  4.    Tiieorem. 

363.  Tlie  area  of  a  parallelogram  is  equal  to  the  prod- 
uct of  its  base  and  altitude. 

Hyp,  Let  ABCD  be  a  CJ,  AB  its  F__D E     ^ 

base,  and  BE  its  altitude.  \   /  \ 

To  prove  ZZ7  ABCD  =  AB  x  BE.  j  /  I  / 

Proof  Draw  AF  J_  to  AB  meet-  t U 

ing  CD  produced  to  F.  ^  ^ 

Then  ABEF  is  a  rectangle  hav- 
ing the  same  base  and  altitude  as  the  dJ  ABCD. 

Then  AD  =  BC,  and  AF  =  BE, 

being  opp.  sides  of  a  CO  (129). 

.-.  A  ADr=  A  BEC, 
Tiaving  the  hypotenuse  and  a  side  respectively  equal  in  each  (110). 

Take  the  A  ADF  from  the  trapezoid  ABCF,  and  there 
is  left  the  z=7  ABCD. 

Take  the  A  BEC  from  the  trapezoid  ABCF,  and  there 
is  left  the  rect.  ABEF. 

r,   OJ  ABCD  =  rect.  ABEF.  (Ax.  3) 

But     area  of  rect.  ABEF  =  AB  x  BE.  (360) 

.  • .  area  of  OJ  ABCD  =  AB  X  BE.  (Ax.  1) 

Q.E.D. 

364.  CoE.  1.  Parallelograms  having  equal  bases  and 
equal  altitudes  are  equivalent,  because  they  are  all  equiva- 
lent to  the  same  rectangle. 

365.  Cor.  2.  Any  two  parallelograms  are  to  each  oilier 
as  the  products  of  their  bases  by  their  altitudes;  therefore 
parallelograms  having  equal  bases  are  to  each  other  as  their 
altitudes,  and  parallelograms  of  equal  altitudes  are  to  each 
other  as  their  bases. 


BOOK  IV.— AREAS  OF  POLYGONS.  185 


Proposition  5.    Theorem. 

366.  Tlie  area  of  a  triangle  is  equal  to  half  the  product 
of  its  base  and  altitude. 

Hyp,  Let  ABC  be  a   A,  BO  its      \; 

base,  and  AD  its  altitude.  \                . 

To  prove  A  ABC  =  I  BC  X  AD.  \          / 

Proof    From   B    draw   BH  ||  to  \     / 

CA,  and  from  A  draw  AH  ||  to  CB  '^/ 

meeting  BH  in  H. 

BCAH  is  a  ZZ7  having  the  base  BC  and  the  altitude  AD.  (124) 

.-.  A  BAC  =  A  BAH, 
a  diagonal  of  a  OO  divides  it  into  two  equal  As  (130). 

But        CJ  BCAH  =  BC  X  AD, 

the  area  of  aOJ  is  equal  to  the  pi'oduct  of  its  base  and  altitude  (363). 

.-.  A  ABC  =  iC7BCAH=:4BC  x  AD.  q.e.d. 

367.  Cor.  1.  Triangles  having  equal  bases  and  equal 
aUiludes  are  equivalent ;  and  therefore  triangles  on  the  same 
base,  and  having  their  vertices  in  the  same  straight  line 
parallel  to  the  base,  are  equivalent, 

368.  Cor.  ^,  If  a  triangle  and  a  parallelogram  have 
the  same  base  and  are  between  the  same  parallels,  the  area 
of  the  triangle  is  half  that  of  the  parallelogram. 

369.  Cor.  3.  Any  two  triangles  are  to  each  other  as  the 
products  of  their  bases  by  their  altitudes ;  therefore  tri- 
angles having  equal  bases  are  to  each  other  as  their  alti- 
tudes, and  triangles  of  equal  altitudes  are  to  each  other  as 
their  bases. 

370.  Cor.  4.  Of  tiuo  parallelogra7ns  or  two  triangles  on 
equal  bases,  that  is  the  greater  which  has  the  greater  alti- 
tude ;  and  of  two  ])arallelograms  or  triangles  of  equal  alti- 
tudes, that  is  the  greater  which  has  the  greater  base. 


186 


PLANE  GEOMETRY. 


Proposition  6.    Theorem. 

371.  The  area  of  a  trapezoid  is  equal  to  the  product  of 
the  half  sum  of  its  parallel  sides  hy  its  altitude. 
Dr, .C 


AH  B 

Hyp,  Let  ABCD  be  a  trapezoid,  AB  and  CD  the  ||  sides, 
and  DH  the  altitude. 

To  prove         area  ABCD  =  i  ( AB  +  CD)  DH. 
Proof    Draw  the  diagonal   BD,  dividing  the  trapezoid 
into  two  AS  ABD  and  DCB,  having  the  common  altitude 
DH. 

Then  area  A  ABD  =  i  AB  x  DH, 

and  area  A  DCB  =  J  CD  X  DH. 

The  area  of  a  A  equals  ^  the  product  of  its  base  and  alt.  (366). 
.• .  area  ABCD  =  J  ( AB  +  CD)  DH, 
since  the  two  as  make  up  the  area  of  the  trapezoid. 

Q.E.D. 

372.  Cor.  Since  the  median  EF  =  i  (AB  +  CD)  (156), 
therefore  the  area  of  a  trapezoid  is  equal  to  the  product  of 
the  median  joining  the  middle  points  of  the  non-parallel 
sides  hy  the  altitude, 

,'.  area  ABCD  :=rE  X  DH. 

373.  ScH.  The  area  of  any  polygon  may  be  found  by 
dividing  it  into  triangles,  and  finding  the  areas  of  the  sev- 
eral triangles.  But  in  practice  the 
method  usually  employed  is  to 
draw  the  longest  diagonal  AF  of 
the  polygon,  and  upon  AF  let  fall 
the  perpendiculars  BM,  CN,  DP, 
EO,  GQ,  thus  decomposing  the 
polygon  into  right  triangles  and 
trapezoids.  By  measuring  the  lengths  of  the  perpendicu- 
lars, and  the  distances  between  their  feet  upon  AF,  the 
areas  of  these  figures  are  readily  found,  and  their  sum  will 
be  the  area  of  the  polygon. 


BOOK  IV.—aOMFAlilSOJSr  OF  AREAS. 


187 


Comparison  of  Areas. 
Proposition  7.    Theorem. 

374.  The  square  described  on  the  hyiMenuse  of  arujlit 
triangle  is  equivalent  to  the  sum  of  the  squares  described 
on  the  other  two  sides, 

Jlyp.  Let  ABC  be  a  rt.  A, 
rt.  angled  at  A,  and  BE,  AK,  AF 
squares  on  BC,  AC,  AB. 

To  2jrove 
sq.  on  BC=sq.  on  AB-fsq.  on  AC. 

Proof  Through  A  draw  AL  ||  to 
BD,  and  join  AD,  FC. 

Since  Z  BAG  is  a  rt.  Z ,  (Cons.) 

and     Z  BAC  is  a  rt.  Z ,    (Hyp.) 

.-.  CAGisast.  line.     (52) 

.  • .  sq.  BG  is  double  the  A  FBC, 
having  ilie  same  base  and  between  the  same  \\  s  (368), 

and        rect.  BL  is  double  the  aABD. 

Again,  since  FB  =  AB,  and  BC  =  BD,         (Cons.) 

and  ZFBC=ZABD, 

eoA^h  being  tlie  sum  of  art.  Z.  and  the  common  Z  ABC, 
.-.  A  FBC=:aABD, 
having  two  sides  and  the  included  Z  equal  each  to  each  (104). 

.-.sq.  BG  =  rect.  BL.  (Ax.  6) 

In  like  manner,  by  joining  BK,  AE,  it  may  be  proved 

thatsq.  HC  =  rect.  CL. 

.  • .  whole  sq.  BDEC  =  sum  of  sqs.  BG  and  HC.  (Ax.  1) 

.  • .  sq.  on  BC  =  sq.  on  AB  -f  sq.  on  AC.     q.e.d. 

Note.— This  proposition  is  commonly  called  the  VytUagorean  Proposition,  be- 
cause it  is  said  to  have  been  discovered  by  Pythagoras  (born  about  600  B.C.). 
Tlie  above  demonstration  of  it  was  given  by  Euclid,  about  300  b.c.  (Prop.  47, 
Book  I.  Euclid). 


188  PLANE  GEOMETRY. 


Proposition  8.    Theorem. 

375.  The  areas  of  two  triangles  having  an  angle  of  the 
one  equal  to  an  angle  of  the  other,  are  to  each  other  as  the 
products  of  the  sides  including  the  equal  angles. 

Hyp.  Let  ABC,  ADE  be  the  two  As 
having  the  common  Z  A. 

A  ABC      AB  X  AC 

To  prove  ^aDE^ADxAE' 

Proof  Join  BE. 

Since  the  As  ABC,  ABE  have  their 
bases  AC,  AE  in  the  same  straight  line, 
and  the  common  vertex  B,  they  have  the  same  altitude. 

A ABC      AC 


aABE      AE' 

AS  of  the  same  altitude  are  to  each  other  as  their  bases  (369). 

Also,  the  A  s  ABE,  ADE  have  the  same  altitude,  since 
their  bases  AB,  AD,  opp.  the  common  vertex  E,  are  in  the 
same  straight  line. 

A ABE      AB 


*  •  A  ADE      AD* 

Multiplying  these  equalities,  we  have 
A  ABC       AB  X  AC 


(369) 


aade-adxae-  ^^^^^ 

Q.E.B. 

376.  Cor.  If  the  products  of  the  sides  including  the 
equal  angles  are  equal,  the  triangles  are  equivalent. 

EXERCISE. 

If  two  triangles  have  an  angle  of  the  one  supplementary 
to  an  angle  of  the  other,  their  areas  are  to  each  other  as  the 
products  of  the  sideB  including  these  angles, 


BOOK  IV.— COMPARISON  OF  AREAS.  189 

Proposition  9.    Theorem. 

377.  The  areas  of  similar  triangles  are  to  each  other 
as  the  squares  of  their  homologous  sides. 


B 
Hyp.  Let  ABC,  A'B'C  be  similar  As. 

A  ABC         BC' 
To  prove  XT^B^  ^  g^,^ ' 

Proof.    Since  Z  B  ==  /  B',  (Hyp.) 

A  ABC    _    BA  X    BC   _  BA        BC^ 

•  •  A  A'B'C  ~  B'A'  X  B'C      B'A'  ^  B'C*       ^  ^^ 

BA         BC 
But 


B'A'      B'C'^ 

homologous  sides  of  similar  as  are  proportional  (307). 

BA  BC 

Substitute  in  the  above  equality  for  ^^^  its  equal  .^v^, 

A  ABC         BC         BC        W 


A  A'B'C  ~  B'C  ^  B'C  ~  JB^/^  *  ^'^'^' 

378.  Cor.  The  areas  of  tivo  similar  triangles  are  to 
each  other  as  the  squares  of  any  ttvo  homologous  lines. 

EXERCISES. 

1.  ABC  is  a  triangle.  AE  and  BF,  intersecting  in  G, 
are  drawn  to  bisect  the  sides  BC,  AC  in  E  and  F.  Com- 
pare the  areas  of  tlie  triangles  AGB,  FGE. 

2.  The  base  and  altitude  of  a  triangle  are  18  and  8  re- 
spectively; wliat  is  the  altitude  of  a  similar  triangle  whose 
base  is  12  ? 

3.  A  has  a  triangular  piece  of  ground,  the  base  of  the 
triangle  being  40  rods;  what  is  the  base  of  a  similarly- 
shaped  lot  containing  4  times  as  much  ?       A71S.  80  rods. 


190 


PLANE  GEOMETRY. 


Proposition  10.    Theorem. 

379.   Tlie  areas  of  similar  polygo7is  are  to  each  other  as 
the  squares  of  their  homologons  sides, 

Hyj).  Let  S  and  S'  denote  the  areas 
of  the  two  similar  polygons  ABODE 
and  A'B'C'D'E',  in  which  AB  and 
A'B'  are  homologous  sides. 


To  prove         ^ 


S       AB 


Proof.  Join  AC,  AD,  A'C,  A'D', 
dividing  the  two  polygons  into  the 
same  number  of  similar  As,  and  simi- 
larly placed.  (321) 

A  ABC 


Also, 


A  A'B'C 
A  ACD 


AB^ 
CD' 


A  A'C'D'      cq5> 


and 


A  ADE 


DE 


A  A'D'E'  ~  5^ 


similar  As  are  to  each  otliei''  as  tlie  squares  of  ilieir  Iwmologous  sides  (377) 
AB  _  M)_  _  OT^  _  ^ 
B'C 


But 


A'B' 
A  ABC 


CD' 
A  ACD 


"  D'E'  ~ 
A  ADE 


etc. 


A  A'B'C  ~  A  A'C'D'  ~  A  A'D'E'* 


(307) 
(Ax.l) 


A  ABC  +  A  ACD  +  A  ADE 
•''a  A'B'C  +  A  A'C'D'  +  A  A'D'E' 


A  ABC 
"aA'B'C 
AB" 


A'B' 


(296) 


A^' 


Q.E.D. 


380.  Cor.  1.   TJie  areas  of  similar  polygons  are  to  each 
other  as  the  squares  of  any  tioo  homologous  lines. 

381.  Cor.  2.   The  homologous  sides  of  similar  polygons 
are  to  each  other  as  the  square  roofs  of  their  areas. 


BOOK  IV.—PBOBLEMS  OF  CONSTRUCTION.        191 

Problems  of  CoNSTRucTioiq". 
Proposition  1  1 .    Problem. 

382.  To  construct  a  triangle  equivalent  to  a  given  poly- 
gon. 

Given,  the  polygon  ABODE. 

Required f  to  construct  a  A  equal 
to  it  in  area. 

Cons.     Join  BD. 

From  C  draw  CF  ||  to  DB,  meet- 
ing AB  produced  in  F ;  join  DF. 

In  like  manner  Join  AD,  draw 
EG  II  to  DA,  meeting  BA  produced  in  G,  and  join  DG. 

Then  area  DGF  =  area  ABODE. 

Proof.  A  BDF  =  A  BDO, 

having  the  same  base  BD  and  their  vertices  G  and  F  in  tlie    ") 
St.  line  GF  \\  to  the  base  (367).  ^ 

Adding  the  polygon  ABDE  to  each  of  these  equalities, 
we  have 

area  AFDE  =  area  ABODE.  (Ax.  2) 

Again,  A  ADG  =  A  ADE. 

Adding  the  A  ADF  to  each,  we  have 

area  GDF  t  area  AFDE  4  area  ABODE. 

Q.E.F. 

383.  ScH.  In  the  same  manner  a  triangle  may  be 
found  equivalent  to  a  polygon  having  any  number  of  sides, 
since  each  operation  of  the  above  process  diminishes  the 
number  of  sides  of  the  polygon  by  one. 


192  PLANE  geometry: 

Proposition  12.    Problem. 

384.  To  construct  a  square  equivalent  to  a  given  paral- 
lelogram. 

Given,  a^ABCD,  CE  its     ^ V  ^ q 

altitude.  |  |    V  jV 

Esquired,   to   construct  a                     I      \  j  \ 

square  of  the  same  area.  I J         V eB 

Cons,     Find  a  mean  pro- 
portional between  AB  and  CE  (349),  and  represent  it  by 
EG. 

The  square  described  upon  the  line  EG  will  be  equiva- 
lent to  the  CJ  ABCD. 

Proof,     Since  AB  :  EG  =  EG  :  CE,  (Cons.) 

.-.  FG'  =  ABx  CE.  (281) 

.• .  area  EGHK  =  area  ABCD.  q.e.f. 

385.  Cor.  1.  To  construct  a  square  equivalent  to  a 
given  triangle,  ice  take  for  its  side  a  mean  j^roportional  be- 
tween the  lase  of  the  triangle  and  one-half  its  altitude*  o^-rx] 

386.  Cor.  2.  A  square  can  ie  found  equivalent  to  any 
given  polygon,  hy  first  constructing  a  tria^igle  equivalent  to 
the  given  polygon  (382),  and  then  constructing  a  square 
equivalent  to  the  triangle  (385). 

Proposition  13.    Problem. 


387.  To  construct  a  square  equivalent  to  the  sum  ofttoo 

P 

Q 


give7i  squares,  q  p 

Given,  P  and  Q  the  sides  of 


the  squares. 

Required,     to     construct      a       ■  = 

square  equivalent  to  their  sum. 


^v^ 


D  E     A  B 


BOOK  IV.— PROBLEMS  OF  CONSTRUCTION.        193 

Cons,  DrawAB  =  P. 

At  A  construct  the  rt.  /_  A,  draw  AC  =  Q,  and  join  BC. 
Construct  the  square  DEFG,  witli  each  of  its  sides  equal 
toBC. 

Then  DEFG  is  the  square  required. 

Proof,  W  =  AB'  +  AC'.  (374) 

.-.  DE'  =  P^  +  Q\  Q.E.F. 

388.  ScH.  In  the  same  wa}^  a  square  may  be  found 
equivalent  to  the  sum  of  any  number  of  squares;  for  the 
same  construction  which  reduces  two  of  them  to  one  will 
reduce  three  of  them  to  two,  and  these  two  to  one,  and  so 
of  others. 


Proposition  14.    Problem. 

389.  To  construct  a  square  equivalent  to  the  difference 
of  ttvo  given  squares. 

Given,  P  and  Q  the  sides  of  p _ 

the  squares.  G  F     Q 

Required,     to     construct    a    f"         ~\      Cjv,^ 
square  equivalent  to  their  dif-  I 

ference.  ! 

Cons.  Draw  the  indefinite  st.    D  e       A  "^B~H 

line  AH. 

At  A  construct  the  rt.  /A,  draw  AC  =  Q,  the  shorter 
side  of  the  given  lines. 

With  centre  C,  and  a  radius  =  P,  describe  an  arc  cutting 
AH  in  B. 

Construct  thq  square  DEFG,  with  each  of  its  sides  equal 
to  AB. 

Then  DEFG  is  the  square  required. 

Proof,  AB'  =  BC'  -  AC'.  (374) 

...   DE':=P-Q\  q.P^.F. 


194  PLANE  GEOMETRY. 

Proposition  15.    Problem. 

390.  To  cQustrtict  a  rectangle  equivalent  io  a  given 
square,  and  having  the  sum  of  two  adjacent  sides  equal  to 
a  give7i  line. 

Given,  the  square  S,  and  the  c     ^^^-^     ^^^^--P 
line  AB  =  the  sum  of  the  sides,     j  /        *  j\ 

Required,  to  construct  a  rect-     [ \\ 

angle  ==  S,  having  the  sum  of   A  E  B 

its  base  and  altitude  =  AB. 

Cons.  Upon  AB  as  a  diameter,  describe  a  semicircle. 

At  A  erect  AC  _L  to  AB  and  =  a  side  of  S. 

Draw  CD  1|  to  AB,  cutting  the  ©ce  at  D,  and  draw 
DE  i,  to  AB. 

Then  AE  and  EB  are  the  base  and  altitude  of  the  re- 
quired rectangle. 

Proof,  AE  X  EB  =  DE', 

mice  DEis  a  mean  proportional  hetioeen  AE  and  EB{d25), 

and  DE  =  AC  =  a  side  of  S.  (Cons.) 

.-.  AE  X  EB  =  S.  Q.E.R 

391.  ScH.  When  the  side  of  the  square  S  exceeds  half 
the  line  AB  the  problem  is  impossible. 

EXERCISES. 

1.  Construct  a  square  equivalent  to  the  sum  of  two  squares 
whose  sides  are  6  and  8  inches.    /  d 

2.  Construct  a  square  equivalent  to  the  difference  of  two 
squares  whose  sides  are  15  and  25  feet.  >  0 

3.  The  perimeter  of  a  rectangle  is  144  feet,  and  the 
length  is  three  times  the  altitude :  find  the  area. 

4.  On  a  given  straight  line  construct  a  triangle  equal  to 
a  given  triangle  and  having  its  vertex  on  a  given  straight 

Mine  not  parallel  to  the  base. 

5.  Construct  a  parallelogram  that  shall  be  equal  in  area 
and  perimeter  to  a  given  triangle.  :^  <7  C  P 


BOOK  IV.— PROBLEMS  OF  GONSTRtfCTtON.        195 


Proposition  1  6.    Problem. 

392.  To  construct  a  rectangle  equivalent  to  a  given 
sqiiare,  and  having  the  difference  of  tioo  adjacent  sides 
equal  to  a  given  line. 

Given,    the   square   S,   and        Cq,^- -^^ 


S 


\ 


the  line  AB  =  the  difference         ;,*^n. 

of  the  sides .  A  \ ^^-s: \  b 

Required,    to    construct    a         \ 
rectangle  =  S,  having  tlie  dif- 
ference of  its  base  and  alti- 
tude =  AB. 

Cons.     Upon  AB  as  a  diameter,  describe  a  O. 

At  A  erect  AC  _L  to  AB  and  =  a  side  of  S. 

Through  C  and  the  centre  of  the  O  draw  CH,  cutting 
the  Oce  at  D  and  H. 

Then  CII  and  CD  are  the  base  and  altitude  of  the  re- 
quired rectangle. 


Proof.  CH  X  CD  =  AC  , 

a  tangent  is  a  mean  proportional  between  tlie  whole  secant  and  ilie 
ext.  seg.  (339). 

.-.CHxCD^S, 

and  the  difference  between  CH  and  CD  is  DH  =  AB. 
.•.  CH  X  CD  is  the  required  rectangle. 

Q.KF. 
EXERCISES. 

1.  The  bases  of  a  trapezoid  are  14  and  16  feet ;  the  non- 
parallel  sides  ai-e  each  6  feet :  find  the  area  of  the  trape- 
zoid. 

2.  Construct  a  rhombus  equal  to  a  given  parallelogram 
and  having  one  of  the  sides  of  the  parallelogram  for  one 
side  of  the  rhombus. 


C            n 

.<: 

1     ^\ 

I 

A           V'" 

B    - 

196  PLANE  GEOMETRY. 


Proposition  1  7.    Problem. 

398.       Tioo  similar  polygons  being  given,  to  construct 
a  similar  polygon  equal  to  their  sum,  p 

Given,  two  homologous  sides  P  and  "^ ^' 

Q  of  two  similar  polygons  R  and  S. 

Required,   to    construct    a  similar^ 
polygon  equivalent  to  their  sum.  Q 

Cons,  Draw  AB  =  P. 

At  A  construct  the  rt.  Z  A,  draw 
AC  =  Q,  and  join  BC.  . 

On  BC,  homologous  to  P  and  Q,  construct  a  polygon  T 
similar  to  R  and  S,  as  in  (354). 

Then  T  is  the  polygon  required. 


Proof  -5  =  ^,and|^^.  (379) 


....  R  +  S        F  +  Q' 

Adding,  —y^  =       ^u   > 

But  BC*  =  P'  +  Q'.  (Cons.) 

.•.T  =  R  +  S.  Q.E.F. 

394.  ScH.  To  construct  a  polygon  similar  to  two  given 
similar  polygons,  and  equivalent  to  their  difference,  we 
find  the  side  of  a  square  equivalent  to  the  difference  of  the 
squares  on  P  and  Q  (389),  and  on  this  side,  homologous  to 
P  and  Q,  construct  a  polygon  similar  to  the  given  polygons 
R  and  S  (354).     This  will  be  the  polygon  required  (379). 


BOOK  IV.-ritOBLEMS  OF  CONSTRUCTION.        197 


Proposition  1 8.    Problem. 


395.  To  find  tivo  straiglit  lines  proportional  to  two  given 

polygons, 

Give7i,  two  polygons  R  and  S.  ^^ 

Required,  to   find  two  st.  lines  ^--^^    j\ 

proportional  to  R  and  S.  /'  |     \ 

Cons.  Find  two  squares  equiva-    .^^. 1. \ 

•lent  to  the  given  polygons  R  and  S    ° 
(386) ;  let  P  and  Q  be  the  sides  of  these  squares. 

Construct  the  rt.  Z  A,  draw  AB  =  V,  and  AC  =  Q. 

Join  BC,  and  draw  AD  _L  to  BC. 

Then  BD  and  DC  are  the  lines  required. 

Proof,    Since  AD  is  a  JL  from  the  rt.  /  A  on  the  hypote- 
nuse BC, 

.-.  AB":  AC'  =  BD:DC, 

tJie  sqs.  of  the  sides  about  ifie  rt.  Z  are  proportional  to  the  adj.  segments 
of  tJie  hypotenuse  (324). 

But  AB  =  P,  and  AC  ^  Q.  (Cons.) 

.•.F:Q'  =  BD:DC, 

and  .  • .  BD,  DO  are  proportional  to  the  areas  of  the  given 
polygons.  Q.E.F. 

EXERCISE. 

Bisect  a  quadrilateral  by  a  straight  line  drawn  fr^m  one 
of  its  vertices. 

Let  ABCD  be  the  quad.;  bisect  BD  in  E,  let  E  lie  between  AC  and  B;  through 
E  draw  EF  ll  to  AC  to  meet  PC  in  F;  join  CE,  EA,  AF,  tljen  AFCP  =  ^APCD, 


p 

— 

r\ 

S 

c 

K               D         B 

"K 

198  PLANE  GEOMETRY. 


Proposition  19.    Problem. 

396.  To  construct  a  square  loliicli  shall  have  to  a  given 
square  the  ratio  of  tioo  given  li7ies. 

Given,  the  square  S,  and  the 
ratio  P  :  Q. 

Required,  to  construct  a  square 
which  shall  be  to  S  as  P  is  to  Q. 

Cons,  Draw  the  st.  line  AK; 
take  AD  =  P  and  DB  =  Q. 

Upon  AB  as  a  diameter,  describe  a  |^Oce. 

At  D  erect  the  _L  DC,  and  join  AC,  BC. 

On  CB,  or  CB  produced,  take  CH  =  a  side  of  S. 

Draw  HEj^  t^BA.  _ 
'"'""Then  CE  is  the  side  of  the  required  square. 

Proof.  Since  CD  is  a  _L  from  the  rt.  /  C  to  the  hypote- 
nuse AB, 

.•.Cr:Cg'  =  AD:DB.  (324 

But  CA  :  CB  =  CE  :  CH, 

a  St.  line  \\io  a  side  of  a  A  cuUilie  other  two  sides  proportionally  (298), 

and  CA'  :  CB'  =  CE'  :  ClI'.  (295) 

.-.  CE'  :  CH'  =  AD  :  DB  =P  :  Q. 
But  CH'  =  S. 

.-.CE^-.S^PlQ.  Q.E.F. 

397.  ScH.  To  construct  a  polygon  similar  to  a  given 
polygon  S,  and  having  to  it  the 
given  ratio  of  P  to  Q,  we  find,  as       /  \ 
in  (396),  a  side  x  so  that  x^  shall     /      ^     / 
be  to  6'  (where  s  is  a  side  of  S)        ~^~  s 
as  P  is  to  Q,  and  upon  x  as  a  side                         q 
homol(f^ous   to  s,  construct  the 

polygon  R  similar  to  S   (354)  j  this  will  be  the  polygon 
required.  (379) 


BOOK  IV,— PROBLEMS  OF  CONSTRUCTION.        199 


Proposition  20.    Problem, 


^ 


398.  To  construct  a  polygon  equivalent  to  a  given  poly- 
gon  F,  and  similar  to  a  given  polygon  Q, 

Given,  two  polygons  P  aud  Q. 

Esquired,  to  construct  a  poly- 
gon equivalent  to  P  and  similar 
toQ.  ^ 

Co7is,  Find  jt?  and  ^,  the  sides   p ^ 

of  squares  equivalent  respectively  /^ 

toPandQ.  (386)  /     \ 

Take  any  side  of  Q  as  AB,  and  \      Q      ^ 

find  a  fourth  proportional  A'B'  to  p -^f 

^,jt?,  andAB.  (347) 

Upon  A'B',  homologous   to  AB,  construct   Q'  similar 
to  Q.  (354) 

Then  Q'  is  the  polygon  required. 

Proof,  Since  Q'  is  similar  to  Q,  (Cons..) 

.•.Q;Q'  =  AB'  :  A7F'\  (379) 

But  q:p  =  AB:  A'B'.  (Cons.) 

.'.(i:q'  =  f:p\  (Ax.  1) 

But  P  =  i/,  and  Q  =  q\  (Cou^.) 

.•.Q:Q'  =  Q:P;  and.-.Q'  =  P. 
.  • .  Q'  is  equivalent  to  P  and  similar  to  Q, 

Q.E.F. 


200  PLANE  GEOMETBY. 

ApPLICATIOI!^S. 

1.  To  find  the  area  of  an  equilateral  triangle  in  terms  of 
its  side  a,* 

Let  h  denote  the  alt.,  and  S  the  area,  of  the  A . 

Then  h  =  Va'  -  T  -  ^  ^3. 

But  (3GG)  S  =  ^^^ . 

2.  To  find  the  area  of  a  triangle  in  terms  of  its  sides  and 
the  radius  of  the  circumscribing  circle. 

Let  a,  h,  c  denote  the  sides  and  h  the  alt.  of  the  A ,  and 
R  the  radius  of  the  circumscribing  O. 
By  Ex.  4  (354),  be  =  2R/^. 

.-.  abc  =  2Jla7i  =  4ES.  (3G6) 

••^-4R- 

Therefore,  tlie  area  of  a  triangle  is  equal  to  the  prodicct 
of  the  three  sides  divided  hy  four  times  the  radius  of  the 
circumscribing  circle. 

3.  To  find  the  area  of  a  triangle  in  terms  of  its  sides. 

By  Ex.  1  (354),  /i  =  -  V s{s  -  a){s  -  b){s  -  c). 
a 

And  since  S  =  ^ah, 

. • .  S  =  Vs{s-a){s-b){s-c), 

5.  Find  the  area  of  an  equilateral  triangle  if  a  side  =  1 
foot.  Ans.  0.433  sq.  ft. 

6.  Find  (1)  the  area  of  the  triangle  whose  sides  are  3,  4, 
and  5  feet,  and  (2)  the  radius  of  the  circumscribing  circle. 

Ans,   (1)  6  sq.  ft.  ;  (2)  2.5  ft. 

*  Rouch^  et  Comberousse,  p.  286. 


BOOK  IV.— EXERCISES.     THEOREMS,  201 


exercises. 
Theorems. 

^        1.  A  triangle  is  divided  by  each  of  its  medians  into  two 
xparts  of  equal  area. 

2.  A  parallelogram  is  divided  by  its  diagonals  into  four 
triangles  of  equal  area. 

3.  ABC  is  a  triangle,  and  its  base  BC  is  bisected  at  D; 
if  H  be  any  point  in  the  median  AD,  show  that  the  tri- 
angles ABH,  ACH  are  equal  in  area. 

4.  In  AC,  a  diagonal  of  the  parallelogram  ABOD,  any 
point  H  is  taken,  and  HB,  HD  are  drawn:  show  that  the 
triangle  BAH  is  equal  to  the  triangle  DAH. 

5.  If  two  triangles  have  two  sides  of  one  respectively 
equal  to  two  sides  of  the  other,  and  the  included  angles 
supplementary,  show  that  the  triangles  are  equal  in  area. 

6.  ABCD  is  a  parallelogram,  and  E,  H  are  the  middle 
points  of  the  sides  AD,  BC ;  if  Z  is  any  point  in  EH,  or 
EH  produced,  show  that  the  triangle  AZB  is  one  quarter 
of  the  parallelogram  ABCD, 

7.  If  ABCD  is  a  parallelogram,  and  E,  H  any  points  in 
DC  and  AD  respectively,  show  that  the  triangles  AEB, 
BHC  are  equal  in  area. 

8.  ABCD  is  a  parallelogram,  and  P  is  any  point  within 
it:  show  that  the  sum  of  the  triangles  PAB,  PCD  is  equal 
to  half  the  parallelogram. 

9.  Find  the  ratio  of  a  rectangle  18  yards  by  14^  yards  to 
a  square  whose  perimeter  is  100  feet. 

10.  The  medians  AD,  BE  of  the  triangle  ABC  are  pro- 
duced to  meet  the  straight  line,  drawn  through  C  parallel 
to  AB,  in  F,  &  respectively:  prove  that  the  triangle  AGC 
is  equal  in  area  to  the  triangle  BCF. 

AD  =  DF  (105),  etc. 

11.  ABCD  is  a  parallelogram ;  through  A  any  two  straight 
lines  AE,  AF  are  drawn  meeting  BC  in  E  and  CD  in  F; 


202  PLANE  GEOMETRY. 

through  D,  DG  is  drawn  parallel  to  AE  meeting  AF  in  G. 
Show  that  the  parallelogram  having  AE,  AG  as  adjacent 
sides  is  equal  to  ABCD. 

Draw  EH  ||  to  AG  meeting  DG  produced  in  H;  produce  BC  to  meet  DH  in  K; 

CO  AH  =  ^Z7  AK;  why?  etc. 

12.  If  0  is  the  centroid  (172)  of  the  triangle  ABC,  prove 
that  the  triangles  AOB,  BOO,  COA  are  equal  in  area. 

13.  Show  that  the  line  joining  the  middle  points  of  the 
parallel  sides  of  a  trapezoid  divides  the  area  into  two  equal 
parts. 

14.  If  any  point  in  one  side  of  a  triangle  be  joined  to  the 
middle  points  of  the  other  sides,  the  area  of  the  quadri- 
lateral thus  formed  is  one-half  that  of  the  triangle. 

15.  Prove  that  any  straight  line  which  bisects  a  parallel- 
ogram must  pass  through  the  intersection  of  its  diagonals. 

Let  EF  bisect  ^Z/ABCD  meeting  AB  in  E,  DC  in  F,  BD  in  O;  join  ED,  FB,  etc. 

16.  APB,  ADQ  are  two  straight  lines  such  that  the  tri- 
angles PAQ,  BAD  are  equal.  If  the  parallelogram  ABCD 
be  completed,  and  BQ  joined  cutting  CD  in  K,  show  that 
CR  =  AP. 

17.  If  the  straight  line  joining  the  middle  points  of  two 
opposite  sides  of  any  quadrilateral  divide  the  area  into  two 
equal  parts,  show  that  the  two  bisected  sides  are  parallel. 

18.  Through  the  vertices  of  a  quadrilateral  straight  lines 
are  drawn  parallel  to  the  diagonals:  prove  that  the  figure 
thus  formed  is  a  parallelogram  which  is  double  the  quadri- 
lateral. 

19.  Through  D,  E  the  middle  points  of  the  sides  BC, 
CA  of  a  triangle  ABC,  any  two  parallel  straight  lines  are 
drawn  meeting  AB  or  AB  produced  in  F  and  G:  prove 
that  the  parallelogram  DEGF  is  half  the  triangle  ABC. 

20.  In  a  trapezoid  the  straight  lines,  drawn  from  the 
middle  point  of  one  of  the  non-parallel  sides  to  the  ends  of 
the  opposite  side,  form  with  that  side  a  triangle  equal  to 
half  the  trapezoid. 

21.  Prove  that  the  points  F,  A,  K^  in  the  figure  of 
Prop.  7,  are  colli  near. 


BOOK  IV,— EXERCISES.     THEOREMS.  203 

IV     22.  In  the  figure  of  Prop.  7,  if  FG,  KH  be  produced  to 
meet  in  P,  prove  that  PA  produced  cuts  BC  at  right  angles. 

23.  Points  E,  F,  G,  II  are  taken  respectively  in  the  sides 
AB,  BC,  CD,  DA  of  a  rectangle  ABCD;  if  EF  =  GH, 
prove  that  

AG'  +  CH'  =  AF'  +  CE'. 

24.  From  D,  the  middle  point  of  the  side  AC  of  a  right 
triangle  ABC,  DE  is  drawn  perpendicular  to  the  hypote- 
nuse AB :  prove  that  BE'  =  AE'  +  BC'. 

25.  ABCD  is  a  parallelogram.  If  AC  is  bisected  in  0 
and  a  straight  line  MON  is  drawn  to  meet  AB,  CD  in  M, 
N  respectively,  and  OR  parallel  to  AB  meets  AN  in  R; 
prove  that  the  triangles  ARM,  CRN  are  equal. 

NC  =  AM  (105),  .-.  AN  is  II  to  CM  (133),  etc. 

26.  If  through  the  vertices  of  a  triangle  ABC  there  be 
drawn  three  parallel  straight  lines  AD,  BE,  CF  to  meet 
the  opposite  side  or  sides  produced  in  D,  E,  F:  prove  that 
the  area  of  the  triangle  DEF  is  double  that  of  ABC. 

Let  D  be  in  BC;  E,  F  in  CA,  BA  produced;  a  EFB  =  a  ECB,  etc. 

27.  ABC,  DEF  are  triangles  having  the  angles  A  and  D 
equal,  and  AB  equal  to  DE :  show  that  the  triangles  are  to 
each  other  as  AC  to  DF. 

28.  The  side  BC  of  the  triangle  ABC  is  bisected  in  D: 
prove  that  any  straight  line  through  D  divides  the  sides 
AB,  AC  into  segments  which  are  proportional. 

29.  On  the  sides  AB,  AC  of  a  triangle  ABC  points  D,  E 
are  taken  such  that  AD  is  to  DB  as  CE  is  to  EA :  if  the 
lines  CD,  BE  intersect  in  F,  prove  that  the  triangle  BFC 
is  equal  to  the  quadrilateral  ADFE. 

30.  ABCD  is  a  square.  A  line  drawn  through  A  cuts 
the  sides  BC,  CD,  produced  if  necessary,  in  E,  F.  Prove 
that  the  triangle  CEF  is  to  the  triangle  ABC  as  the  differ- 
ence of  the  lines  CE  and  CF  is  to  BC. 

31.  0  is  a  point  inside  a  triangle  ABC;  AO,  BO,  CO 
produced  meet  the  sides  in  D,  E,  F,  respectively.  If  AO  is 
to  OD  as  BO  to  EO  as  CO  to  OF,  prove  that  0  is  the  cen- 
troid  (172)  of  the  triangle  ABC. 


204  PLANE  OEOMETRT. 

32.  ABO  is  a  triangle,  and  points  D,  E  are  taken  in  BC, 
CA  respectively,  such  that  BD  =  4  DC  and  AE  =  ^AC; 
AD,  BE  intersect  in  0.     Prove  that  A  AOB  =  i  A  ABC. 

A  BAD  :  A  ADC  =  BD  :  DC  =  1  :  2;  sim.  a  OBD  :  a  ODC  =1:2,  etc. 

33.  ABCD  is  a  quadrilateral  having  two  of  its  sides  AB, 
CD  parallel;  AF,  CG  are  drawn  parallel  to  each  other 
meeting  BC,  AD  respectively  in  F,  G.  Prove  that  BG^  is 
parallel  to  DF. 

Produce  CB,  DA  to  meet  in  E,  etc. 

34.  ABC  is  a  triangle,  G  its  centroid,  D  the  middle  point 
of  BC,  AE  its  perpendicular  to  BC.  Show  that  if  the  rect- 
angle of  which  AE,  EC  are  adjacent  sides  be  completed, 
the  fourth  corner  being  F,  then  FG  produced  bisects  BE. 

Produce  FG  to  meet  BC  in  H;  HD  :  AF  =  1  :  2,  etc. 

35.  ABCD,  AB'C'D'  are  two  squares,  BAB',  DAD'  being 
straight  lines  ;  B'C  meets  AD  in  E,  and  CD  meets  AB'  in 
F.     Prove  that  AE  ^AF. 

36.  A  triangle  ABC  has  the  side  AB  =  2AC  ;  from  C  is 
drawn  CD  to  a  point  D  in  AB  such  that/ ACD  =  Z  ABC: 
show  that  ABCD  ==  3aACD. 

37.  If  in  an  isosceles  triangle  ABC  the  two  equal  sides 
AB,  CA  be  divided  at  F  and  E  respectively  in  any  given 
ratio ;  the  straight  line  FE  will  meet  BC  produced  at  a 
point  D,  such  that  CD  :  BD  =  AF'  :  FB',  or  =  CE'  :  EA'- 

Draw  AG  II  to  CB  meeting  EF  produced  in  G,  etc. 

38.  ABC  is  an  isosceles  triangle  having  the  sides  AB,  AC 
equal,  and  is  such  that  if  BD  be  drawn  bisecting  the  angle 
ABC  and  cutting  AC  in  D,  then  AD  is  equal  to  BC.  Show 
that  me  angle  C  is  double  the  angle  A. 

39.  Perpendiculars  are  drawn  from  the  vertices  of  a  tri- 
angle on  any  straight  line  through  the  centroid  of  the 
triangle  :  prove  that  one  of  these  perpendiculars  is  equal 
to  the  sum  of  the  other  two. 

40.  Perpendiculars  are  drawn  from  the  vertices  and  the 
centroid  of  a  triangle  on  a  given  straight  line  :  prove  that 
the  perpendicular  from  the  centroid  is  the  arithmetic  mean 
of  the  perpendiculars  from  the  vertices. 

Let  ABC  be  the  a  ,  O  its  centroid,  D  tiie  mid.  pt.  of  BC;  draw  AL,  BM,  CN,  DR, 
OK  ±  to  given  line;  bisect  AO  in  E;  draw  EH  j.  to  LM,  etc. 


BOOK  IV.— EXERCISES.     THEOREMS. 


205 


41.  If  the  vertical  angle  C  of  a  triangle  ABC  be  bisected 
by  a  line  which  meets  the  base  in  D,  and  is  produced  to  a 
point  E,  so  that  the  rectangle  of  CD  and  CE  is  equal  to 
that  of  AC  and  CB;  show  that  if  the  base  and  vertical 
angle  be  constant  the  position  of  E  is  fixed. 

Since  CD  x  CE  =  ACxCB,.-.CD  :  AG  =  BC:  C£,and  Z  ACD  =  zBCD, .'.  ZCAB 
=  ZCEB,  etc. 

42.  ABC  is  a  triangle  inscribed  in  a  circle;  from  A 
straight  lines  AD,  AE  are  drawn  parallel  to  the  tangents 
at  B,  C  respectively,  meeting  BC,  produced  if  necessary,  in 
D,  E:  prove  that  BD  is  to  CE  as  the  square  on  AB  is  to 
the  square  on  AC. 

Let  BF,  CG  be  the  tangs, :  ZADB  =  alt.  /DBF  =  ZBAC  (238  and  243),  and  ZB 
is  common  to  both  as  BAC,  BAD,  etc. 

43.  If  P  be  a  point  on  a  diameter  AB  of  a  circle,  and  PT 
be  the  perpendicular  on  the  tangent  at  a  point  Q  :  show 
that  PT  X  AB  =  AP  X  PB  +  PQ'. 

Produce  QP  to  meet  the  0  in  R;  draw  diam.  RS:   Z  TQP  =  ZQSR,  etc. 

44.  Prove  that  the  square  constructed 
on  the  sum  of  two  straight  lines  is  equiva" 
lent  to  Ihe  sum  of  the  squares  con- 
structed on  the  two  lines,  together  with 
twice  the  rectangle  of  the  lines. 

45.  Prove  that  the  square  constructed 
on  the  difference  of  two  straight  lines 
is  equivalent  to  the  sum  of  the  squares 
constructed  on  the  lines,  diminished  by 
twice  the  rectangle  of  the  lines. 

46.  Prove  that  the  rectangle  of  the 
sum  and  the  difference  of  two  straight 
lines  is  equivalent  to  the  difference  of 
the  squares  constructed  on  the  lines, 

47.  ABC  is  an  isosceles  triangle,  CA,  CB  being  the  equal 
sides;  BO  is  drawn  at  right  angles  to  BC  to  meet  CA  pro- 
duced in  0:  show  that  the  square  on  OB  is  equal  to  the 
square  on  OA  with  twice  the  rectangle  OA,  AC. . 

48.  If  a  straight  line  be  divided  into  two  equal  and 


D   G 


H 


D'    E     D 


206  PLANE  GEOMETRY. 

also  into  two  unequal  parts,  the  rectangle  of  the  unequal 
parts  with  the  square  on  the  line  between  the  points  of 
section,  is  equal  to  the  square  on  half  the  line.  (Euclid,  B. 
II,  prop.  5.) 

49.  The  square  on  the  straight  line,  drawn  from  the  ver- 
tex of  an  isosceles  triangle  to  any  point  in  the  base,  is  less 
than  the  square  on  a  side  of  the  triangle  by  the  rectangle  of 
the  segments  of  the  base. 

50.  The  sides  AB,  CD  of  a  quadrilateral  ABCD  inscribed 
in  a  circle  are  produced  to  meet  in  P;  and  PE,  PF  are 
drawn  perpendicular  to  AD,  BO  respectively:  prove  that 
AE  is  to  ED  as  CF  to  FB. 

51.  The  sides  AB,  AD,  produced  if  necessary,  of  a  par- 
allelogram ABCD,  meet  a  line  through  C  in  E,  F  respect- 
ively; CB,  CD,  produced  if  necessary,  meet  a  line  through 
A  in  G,  H  respectively.     Prove  that  GE  is  parallel  to  HF. 

Problems. 

52.  Construct  a  square  equal  to  three  given  squares. 

53.  Construct  a  square  which  shall  be  five  times"  as  great 
as  a  given  square. 

54.  Construct  a  triangle  equal  in  area  to  a  given  quadri- 
lateral. 

55.  Construct  an  isosceles  triangle  equal  in  area  to  a 
given  triangle  and  having  a  given  vertical  angle. 

56.  Divide  a  straight  line  into  two  parts,  so  that  the  sum 
of  the  squares  on  the  parts  may  be  equal  to  a  given  square. 

57.  Trisect  a  triangle  by  straight  lines  drawn  from  a 
given  point  in  one  of  its  sides. 

68.  Find  a  point  0  inside  a  triangle  ABC  such  that  the 
triangles  OAB,  OBC,  OCA  are  equal. 

59.  Construct  a  square  that  shall  be  one-third  of  a  given 
square. 

60.  Divide  a  straight  line  into  two  parts  so  that  the  rect- 
angle contained  by  the  whole  and  one  part  may  be  equal  to 
a  given  square.  

Let  AB  be  the  given  line.    Draw  BD  ±  to  A.B  so  that  BD"  =  given  square; 
join  AD,  draw  DE  ±  to  AD,  meeting  AB  produged  in  E. 


BOOK  IV.— EXERCISES.— PROBLEMS.  207 

61.  Produce  AB  to  0  so  that  the  rectangle  of  AB  and 
AC  may  be  equal  to  a  given  square. 

62.  Construct  a  parallelogram  equal  to  a  given  triangle 
and  having  one  of  its  angles  equal  to  a  given  angle. 

63.  Given  three  similar  triangles:  construct  another  sim- 
ilar triangle  and  equivalent  to  their  sum.  (377.) 

64.  Construct  a  triangle  similar  to  a  given  triangle  ABC 
which  shall  be  to  ABC  in  the  ratio  of  AB  to  BC. 

Find  DE  a  mean  proportional  between  AB,  BC,  etc. 

65.  Bisect  a  triangle  by  a  straight  line  parallel  to  one  of 

its  sides. 

Bisect  AC  of  the  a  ABC  in  D;  from  AC  cut  off  A£  a  mean  proportional  to 
AC,  AD,  etc. 

66.  A  is  a  point  on  a  given  circle:  draw  through  A  a 
straight  line  PAQ  meeting  the  circle  in  P  and  a  given 
straight  line  in  Q,  so  that  the  ratio  of  PA  to  AQ  may  be 
equal  to  a  given  ratio. 

Draw  AB  to  any  pt.  B  in  the  given  line  MN;   produce  BA  to  C   so  that 
BA  :  AC  =  given  ratio,  etc. 

67.  From  the  vertex  of  a  triangle  draw  a  line  to  the  base, 
so  that  it  may  be  a  mean  proportional  between  the  segments 
of  the  base. 

About  the  given  a  describe  a  © ,  etc. 

68.  Show  how  to  draw  through  a  given  point  in  a  side 
of  a  triangle  a  straight  line  dividing  the  triangle  in  a  given 
ratio. 

69.  Construct  a  triangle  equal  to  a  given  triangle  and 
having  one  of  its  angles  equal  to  an  angle  of  the  triangle, 
and  the  sides  containing  this  angle  in  a  given  ratio. 

70.  Draw  through  a  given  point  a  straight  line,  so  that 
the  part  of  it  intercepted  between  a  given  straight  line  and 
a  given  circle  may  be  divided  at  the  given  point  in  a  given 
ratio.  *' 

Let  P  be  the  given  pt,,  XY  the  given  line,  A  the  cent,  of  given  0 :  produce  AP 
to  meet  XY  in  Q;  in  PA  take  B  so  that  QP :  PB  ==  given  ratio,  etc. 


Book  V. 


EEGULAR  POLYGONS.    THE  CIRCLE. 
MAXIMA  AND  MINIMA. 


Regular  Polygons. 

399.  Def.  a  regular  polygon  is  a  polygon  which  is  both 
equilateral  and  equiangular;  as,  for  example,  the  equilateral 
triangle  and  the  square. 

Proposition   1 .    Theorem. 

400.  If  the  circumference  of  a  circle  he  divided  into  any 
numher  of  equal  arcs,  (1)  the  cliords  of  these  arcs  form  a 
regular  polygon  inscribed  in  the  circle,  and  (2)  the  tan- 
ge?its  at  the  points  of  division  form  a  regular  polygon  cir- 
cumscribed about  the  circle. 

Hyp.  Let  the  Oce  be  divided  into 
equal  arcs  at  the  pts.  A,  B,  C,  etc.; 
let  AB,  BO,  CD,  etc.,  be  chords  of 
these  arcs,  and  GBH,  HCK,  etc.,  be 
tangents. 

(1)  To  prove  that  ABCD is  a 

regular  polygon. 


BC  =  arc   CD  =  etc.,  (Hyp.) 
BC  =  chd.CD  =  etc., 


Proof  Since  arc  AB  =  arc 
.-.  chd.  AB  =  chd 
in  ilie  same  O  equal  arcs  are  subtended  by  equal  cliords  (194), 
and  ZFAB=:    ZABC=    /  BCD  =  etc., 

all  As  inscribed  in  equal  segments  are  equal  (239). 

.  • .  ABCDEF  is  a  regular  polygon.  (399) 

(2)  To  prove  that  GHK  ...  is  a  regular  polygon. 
Proof  In  the  As  ABG,  BCH,  etc., 

AB  =  BC  =  CD  =  etc., 
and     zCABr=zGBA  =  zHBC=ZHCB=etc., 
being  measured  by  halves  of  equal  arcs  (243). 
.  • .  the  As  ABG,  BCH^  etc.,  are  all  equal  and  isosceles. 

208 


BOOK  V.—REGVLAll  POLYGONS.  209 

.-.  AG  =  BG  =  BE  =  CH  =  etc., 

/G  =  zH=ZK  =  etc.(    li.  i^S    ] 

.-.  GH  =  IIK  =  etc.,  --kr^tLSf^*   ^) 

and         .*.  GHKL  ...  is  a  regular  polygon.  (399) 

Q.E.D. 

401.  Cor.  1.  If  a  regular  inscribed  'polygon  is  given, 
the  tangents  at  the  vertices  of  the  given  polygon  form  a 
regular  civ ciimscrihcd  polygon  of  the  same  numler  of  sides, 

402.  Cor.    2.    //"  a  regular  i7i-  .r     ^^     ^r 
scribed  polygon  ABCD , . .  .  is  given,          /v^  j  ^^ 
the  tangents  at  the  middle  poi7its  M,      /y     \  I  /^  \\^ 
N,  P,  etc.,  of  the  arcs  AB,  BG,  CD,    A/       M^"^^  ^^   . 
etc.,  form   a   regular  circumscribed    nK^^        /^\\9/ 
polygon  ichose   sides  are  parallel   to     V\   /     \    /jp 
those  of  the  i^iscribed  polygon,  and         yC^     ^^X/^ 
tvhose  vertices  A',  B' ,  C",  etc,  lie  on  ^ 

the  radii  OAA',  OBB',  etc  For,  the  sides  AB,  A'B' 
are  || ,  being  _L  to  OM,  (204)  and  (210),  and  the  same  for 
the  others ;  also,  since  B'M  =  B'N  (268),  the  pt.  B'  must 
lie  on  the  bisector  OB  (160)  of  the  Z  MON. 

403.  Cor.  3.  If  the  chords  AM,  MB,  BN,  etc.,  be 
draion,  the  chords  form  a  regular  inscribed  polygon  of 
double  the  number  of  sides  of  ABCD  .... 

404.  Cor.  4.  If  through  the  poinds  A,  B,  C,  etc.,  tan- 
gents  are  draiun  intersecting  the  tangents  A' B' ,  B'C,  etc,  a 
regular  circumscribed  polygon  is  formed  of  double  the 
munber  of  sides  q/' A'B'C'D' 

405.  ScH.  'It  IS  clear  that  the  area  of  any  inscribed 
polygon  is  less  than  the  area  of  the  inscribed  polygon  of 
double  the  number  of  sides  ;  and  the  area  of  a  circum- 
scribed polygon  is  greater  than  that  of  the  circumscribed 
polygon  of  double  the  number  of  sides. 


210  PLANE  OEOMETBT, 


Proposition  2.    Theorem. 

406.  A  circle  may  he  circumscribed  about  a  regular 
poly g 071,  and  a  circle  may  be  inscribed  in  it. 

Hyp,     Let  ABCDEF  be  a  regular  polygon. 

(1)  To  prove  that  a  O  may  be  circumscribed  about  it. 

Proof,  Bisect  the  /  s  A  and  B^  and  let  the  bisectors 
meet  at  0. 


Since  zFAB=ZABa  (Hyp.) 

.-.  Z0AB=  ZOBA,  (Ax.  7) 

and  .  • .  OA  =  OB.      (jyMj^  /\] 

Join  00,  OD. 

In  the  AsOAB  and  OBO, 

AB  =  BO  (399),  OB  is  common, 

and  Z  OBA  =  Z  OBO.  (Cons.) 

.-.  aOBA=  a  OBO.  (104) 

.-.  Z0AB=  zOCB  =  i  ZBOD, 

and  OA  =  00. 

In  the  same  way  it  may  be  shown  that  each  Z  of  the 


BOOK  v.— REGULAR  POLYGONS.  211 

polygon  is  bisected  by  the  line  joining  it  with  the  pt.  0, 
and  that  OA  =  OB  =  OC  =  OD  =  etc. 

.  • .  the  O  described  with  centre  0  and  radius  OA  will 
pass  through  each  of  the  angular  points. 

(2)   To  prove  that  a  O  may  be  inscribed  in  ABCDEF. 

Proof,  Since  the  sides  AB,  BC,  etc.,  are  equal  chords 
of  the  circumscribed  circle,  they  are  equally  distant  from 
the  centre.  (206) 

Therefore,  if  a  circle  be  described  with  the  centre  0  and 
the  perpendicular  OH  as  a  radius,  this  circle  will  be  in- 
scribed in  the  polygon.  q.e.d. 

407.  Defs.  The  point  0,  which  is  the  comiWn  centre 
of  the  inscribed  and  circumscribed  circles,  is  callM^the 
centre  of  the  regular  polygon. 

The  radius  of  a  regular  polygon  is  the  radius  OA  of  the 
circumscribed  circle. 

The  apothem  of  a  regular  polygon  is  the  radius  OH  of 
the  inscribed  circle. 

The  aiigle  at  the  centre  of  a  regular  polygon  is  the  angle 
between  tw^o  radii  drawn  to  the  extremities  of  any  side,  as 
AOB. 

408.  CoK.  1.  The  inscribed  and  circumscribed  circles 
of  a  regular  polygon  are  concentric, 

/  409.  Cor.  2.  The  7)erpendicular  bisectors  of  the  sides  of 
(I  regnhir  pohjgoii  (ill  jxiss  iJiroinjli  it^i  ccjitre, 

^  410.  Cor.  3.  Tlie  radiu,^  draiun  to  any  vertex  of  a 
regular  polygon  bisects  the  angle  at  the  vertex, 

411.  Cor.  4.  If  liyies  be  draiun  from  the  ceiitre  of  a 
regular  polygon  to  each  of  its  vertices,  the  polygon  will  be 
divided  into  as  m-any  equal  isosceles  triangles  as  it  has 
sides. 

412.  Cor.  ^.  Each  angle  at  the  centre  of  a  regular 
polygon  is  equal  to  four  right  angles  divided  by  the  member 
of  sides  of  the  polygon. 

413.  Cor.  6.  The  interior  angle  of  a  regular  polygon  is 
tlw  supplement  of  the  angle  at  the  centre. 


212  PLANE  GEOMETRY. 


Proposition  3.    Theorem. 

414.  Regular  ])oly (J ons  of  the  same  7iumber  of  sides  are 
similar. 

Hyp.  Let  P  and  P'  be  two 
regular  polygons  of  the  same 
number  of  sides.  p^ 

To  prove  that  P  and  P' 
are  similar. 

Proof  Since  the  polygons 
are  regular, 

.-.  AB   =  BC   =  CD   =etc., 

and  A'B'  =  B'C  =  CD'  =  etc. 

AB   _    B^  _  ^  _ 

•'•A'B'""  B'C'~C'D'~ 

Also,  if  the  polygons  have  each  n  sides,  the  sum  of  all 
the  int.  Zs  of  each  polygon  is  (2;^  —  4)  rt.  Zs.  (148) 

Since  the  polygons  are  equiangular  (399),  and  have  each 

the  same  numbel-  of  sides,  (Hyp.) 

2^j 4 

.  • .  each  Z  of  each  polygon  = —  rt.  Z  s.     (150) 

.-.  ZA=  ZA',   ZB::^  zB',   ZC=  ZC,  etc. 
Hence  the  polygons  are  mutually  equiangular  and  have 
their  homologous*  sides  proportional. 

.  • .  the  polygons  are  similar.  (307) 

Q.E.D. 

415.  Cor.  1.  Tlie  perimeters  of  regular  polygons  of  the 
same  number  of  sides  are  to  each  other  as  any  two  homolo- 
(jous*  sides.  (322) 

416.  Cor.  2.  The  ai^eas  of  regular  polygons  of  the  same 
nmnher  of  sides  are  to  each  other  as  the  squares  of  any  two 
homologous"^  sides.  (379) 

*  Since  the  polygons  are  regular,  any  side  of  one  may  be  taken  as  homologous 
to  any  side  of  the  other. 


BOOK  V.-REGULAR  POLYGONS.  ^13 


Proposition  4.    Theorem, 

417.  Tlie  perimeters  of  any  two  regular  polygons  of  the 
same  number  of  sides  are  to  each  other  as  the  radii  of  their 
circumscribed  circles,  or  as  the  radii  of  their  inscribed 
circles. 

Hyp.  Let  P  and  P'  be  a  H   B  a'    h'  B^ 

the  perimeters  of  two  regn- 
iilar  polygons,  0  and  0' 
their  centres,  OA  and  O'A' 
the  radii  of  their  circum- 
scribed OS,  OH  and  O'H' 
the  radii  of  their  inscibed  ©s. 


To  prove 

P        OA         OH 
P'  ~  O'A'  ~  O'H' 

Proof 

P        AB 
P'-  A'B'- 

In  the  isosceles  As  OAB,  O'A'B', 

ZAOB=zA'0'B'. 

(415) 


(412) 

/ .  the  isosceles  A  s  OAB,  O'A'B'  are  similar,      (309) 

and  .-.  the  rt.  As  OAH,  O'A'H'  are  similar,    (309) 

AB  _   OA  ,   OA  _  OH 

*  •  A'B'  -  O'A' '  ^''"^  O'A'  "  O'H'  •  ^^^  ^^ 

.     P  _J0A  __0H 
•  •  P'  ~  O'A'  ""  O'H'  •  ^        ^ 

Q.E.D. 

418.  Cor.  Tlie  areas  of  regular  polygons  of  the  same 
number  of  sides  are  to  each  other  as  the  squares  of  the  radii 
of  their  circu^nscribed,  or  of  their  inscribed  circles,      (380) 


214  PLANE  GEOMETRY. 


Proposition  5.    Theorem. 

419.  The  area  of  a  regular  polygon  is  equal  to  half  the 
product  of  its  perimeter  and  apothem. 


Hyp.  Let  P  denote  the  perimeter  and  R  the  apothem 
OH,  of  the  regular  polygon  ABCDEF. 

To  prove  area  ABCDEF  =  J  P  X  R. 

Proof  Join  OA,  OB,  OC,  etc. 

The  polygon  is  divided  into  equal  as  whose  bases  are 
the  equal  sides  of  the  polygon  and  whose  common  altitude 
is  the  apothem. 

Then,  area  OAB  =  i  AB  X  OH, 

the  area  ofaA=i  the  product  of  its  base  and  altitude  (366). 

Similarly,        area  OBC  =  ^  BC  X  OH, 

and  so  on  for  all  the  A  s  of  the  polygon. 

.  • .  area  OAB  +  area  OBC  +  etc.  =  4(AB  +  BC  +  etc.)  OH. 

.  • .  area  of  the  polygon  ABCDEF  =  |  P  x  R.  Q.  e.  d. 

420.  CoK.  The  area  of  any  polygon  that  circumscrihes  a 
circle  is  equal  to  half  the  product  of  its  peri7neter  and  the 
radius  of  the  circle. 


BOOK  v.— THE  MEASURE  OF  THE  CIRCLE.        215 


The  Measure  of  the  Circle. 

421.  Def.  a  variable  quantity,  or  simply  a  variable,  is 
a  quantity  which  may  have  an  indefinite  number  of  differ- 
ent successive  values. 

A  constaiit  is  a  quantity  whose  value  does  not  change 
in  the  same  discussion. 

422.  Limit.  When  the  successive  values  of  a  variable, 
under  the  conditions  imposed  upon  it,  approach  more  and 
more  nearly  to  the  value  of  some  constant  quantity,  which 
it  can  never  equal,  yet  from  which  it  may  be  made  to  differ 
by  as  small  as  we  please,  the  constant  is  called  the  limit  of 
the  variable ;  the  variable  is  said  to  approach  indefinilely 
to  its  limit. 

423.  For  example,  suppose  C         D    E 

a  point  to  start  at  A  and  move    a  B 

along  AB  towards  B  under  the 

condition  that,  during  successive  seconds  of  time,  the  point 
moves  first  half  the  distance  from  A  to  B,  that  is  to  C  ; 
then  half  the  remaining  distance,  or  to  I),  then  half  the 
remaining  distance,  or  to  E,  and  so  on  indefinitely.  Then 
the  distance  from  A  to  the  moving  point  is  a  variable  whose 
limit  is  the  distance  AB.  For,  however  long  the  point 
may  move,  under  these  conditions,  there  will  always  remain 
some  distance  between  it  and  the  point  B,  so  that  the 
distance  from  A  to  the  moving  point  can  never  equal  AB  ; 
but  as  the  moving  point  can  be  brought  as  near  as  we  please 
to  B,  its  distance  from  A  can  be  made  to  differ  from  the 
distance  AB  by  as  little  as  we  please. 

If  we  call  the  distance  AB  2,  then  the  distance  the  point 
moves  the  first  second  will  be  1,  the  distance  moved  the 
next  second  will  be  J,  the  distance  moved  the  third  second 
will  be  J ,  and  so  on.     Therefore  the  whole  distance  from  A 


216  PLANE  OEOMETRY.    - 

to  the  moving  point  tit  the  end  of  n  seconds  is  the  sum  of 
n  terms  of  the  series 

1  +  i  +  i+i  +  etc. 

Now  it  is  evident  that,  however  many  terms  of  this  series 
are  taken,  the  sum  can  never  equal  2;  but  by  taking  a  great 
number  of  terms  the  sum  can  be  made  to  diifer  from  2  by 
as  little  as  we  please.  Hence  we  say  that  the  limit  of  the 
sum  of  the  series  as  the  number  of  terms  is  indefinitly  in- 
creased is  2. 

The  limit  of  the  fraction  1,  as  x  is  indefinitely  increased, 
is  zero  ;  for,  by  increasing  x  at  pleasure,  ?  may  be  made  to 
approach  as  near  as  we  please  to  the  value  zero,  but  can 
never  be  made  exactly  equal  to  zero. 

424.  Principle  of  Limits.  Theorem.  If  two  variables  are 
always  equal  and  each  approaches  a  limit,  the  tioo  limits 
inust  be  equal. 

For,  two  variables  that  are  always  equal  have  always  the 
same  common  value,  that  is,  they  are  really  but  a  single 
variable;  and  it  is  clear  that  a  single  variable  cannot  at  the 
same  time  approach  indefinitely  to  two  unequal  limits. 

425.  Cor.  If  two  variables,  while  approaching  their 
resjjective  limits,  are  always  in  the  same  ratio,  their  limits 
are  m  the  same  ratio. 

For,  if  X  and  y  are  two  variables  in  the  constant  ratio  m, 
so  that  X  =  my,  then  the  two  variables  x  and  my  are  always 
equal  to  each  other,  and  their  limits  are  equal  (424);  there- 
fore, if  a  and  b  are  the  limits  of  x  and  y  respectively,  we 
have  a  =  mb;  that  is,  a  and  b  are  in  the  same  constant 
ratio  m, 

EXERCISE. 

The  apothem  of  a  regular  pentagon  is  3  and  a  side  is  2: 
find  the  perimeter  and  area  of  a  regular  pentagon  whose 
apothem  is  G. 


BOOK  v.— TUB  MEASURE  OF  TEE  CIRCLE,       217 


Proposition  6.    Theorem. 

426.  Every  coiivex  curve  is  less  than  any  enveloping 
line  whatever  that  has  the  same  extremities* 

Hyp.  Let  ABO  be  a  convex  curve, 
and  ADIIC  any  line  enveloping  it  and 
terminating  at  A  and  0. 

To  prove  that  ABC  <  ADHC. 

Proof.  Of  all  the  lines  enveloping 
the  area  ABC,  there  must  be  at  least 
one  line  shorter  than  any  other. 

Now  ADHC  cannot  be  this  line. 

For,  draw  the  tangent  AH  to  the  curve  ABC. 

Then,  the  line  AHC  <  the  line  ADIIC, 

dnce  the  st.  line  AH  is  <  the  curve  ADH  (Ax.  10). 

.  • .  ADHC  is  not  the  shortest  line. 

In  the  same  way  it  may  be  shown  that  no  other  envelop- 
ing line  can  be  the  shortest. 

.  • .  the  curve  ABC  is  less  than  any  enveloping  line.  Q  e.d. 

427.  Cor.  1.  The  circumference  of  a  circle  is  less  titan 
the  perimeter  of  any  circumscribed  polygon,  and  greater 
than  the  perimeter  of  any  inscribed  polygon, 

428.  Cor.  2.  The  perimeter  of  a  regular  inscribed 
polygo7i  is  less 'than  the  perimeter  of  a  regular  inscribed 
polygon  of  double  the  number  of  sides, 

429.  Cor.  3.  The  perimeter  of  a  regular  circum,scribed 
polygon  is  greater  than  the  perimeter  of  a  regular  circum- 
scribed polygon  of  do\ible  the  number  of  sides. 


^18 


PLANE  GEOMETRT. 


Proposition  7.    Theorem. 

430.  If  a  regular  jiolyg on  he  inscribed  in  a  given  circle, 
and  another  regular  polygon  he  circumscribed  about  the 
same  circle,  and  if  the  number  of  sides  of  the  polygons  be  in- 
creased indefinitely,  then  the  perimeter  of  each  of  the  poly- 
gons approaches  the  circumference  of  the  given  circle  as  its 
limit,  and  the  area  of  each  approaches  the  area  of  the  circle 
as  its  limit. 

Hyp,  Let  0  be  the  centre  of  the 
given  O,  and  AB,  CD  homologous  sides 
of  the  regular  inscribed  and  circum- 
scribed polygons  of  the  same  number 
of  sides ;  let  ;;  and  P  denote  the  pe- 
rimeters, s  and  S  the  areas  of  the  in- 
scribed and  circumscribed  polygons 
respectively. 

To  prove  that  the  limit  of  p  and  P  is  the  Oce  of  the  O, 

and  the  limit  of  s  and  S  is  the  area  of  the  O. 

Proof  Join  OF  and  00. 

Then,  since  the  polygons  are  similar, 

,\V'.p  =  OQ'.0¥,  (417) 


and 


S : 5  =  OC   :  OF 


(418) 


I^ow  let  the  number  of  sides  be  increased  indefinitely,  j 

always  remaining  the  same  in  each  polygon;  then  the  angle  j 

FOC  will  diminish  indefinitely,  and  the  pt.  C  will  approach  j 

the  pt.  F,  as  near  as  we  please.  J 

.  • .  DC  will  approach  OF  as  its  limit, 
and  OC"  will  approach  OF'^  as  its  limit. 


BOOK  v.— THE  MEASURE  OF  THE  CIRCLE.        219 

.*.  ultimately  P  andju,  and  also  S  and  s,  will  differ  from 
each  other  by  less  than  any  assignable  quantity. 

.'.  the  Oce  of  the  O,  whjcliJ.s._always  iirtjQrmediate  be;: 
tween  P  and  p  (427),  is  the  limit  of  either  of  the  perime- 
ters ; 

and  the  area  of  the  O,  which  is  clearly  always  interme- 
diate between  S  and  s,  is  the  limit  of  either  of  the  areas. 

Q.E.D. 

431.  Cor.  When  the  number  of  sides  of  the  polygons  is 
increased  indefinitely,  the  radius  of  the  circle  is  the  limit 
of  the  radius  of  the  circumscribed  polygon  and  the  apothem 
of  the  inscribed  polygon, 

432.  Def.  Similar  arcs,  sectors,  and  segments  are  those 
which  correspond  to  equal  angles  at  the  centre,  in  circles 
of  different  radii. 

EXERCISES. 

1.  The  apothem  of  a  regular  pentagon  is  G  and  a  side  is 
4:  find  the  perimeter  and  area  of  a  regular  pentagon  whose 
apothem  is  8. 

2.  If  a  point  be  taken  within  a  regular  polygon,  prove 
that  the  sum  of  its  distances  from  the  sides  of  the  pol3^gon 
is  equal  to  the  radius  of  the  inscribed  circle  multiplied  by 
the  number  of  sides  of  the  polygon. 

3.  ABODE  is  a  regular  pentagon;  AC  and  BE  are  joined 
so  as  to  intersect  in  F:  prove  that  CDEF  is  a  parallelo- 
gram. 

4.  The  radius  of  a  circle  is  8 :  find  the  apothem,  perimeter, 
and  area  of  the  inscribed  equilateral  triangle. 

5.  The  radius  of  a  circle  is  10:  find  the  perimeter  and 
area  of  the  regular  inscribed  octagon. 

6.  The  radius  of  a  circle  is  4:  find  the  area  of  the  in- 
scribed square. 


1 


220 


PLANE  GEOMETRY. 


Proposition  8.    Theorem. 

433.  The  circumferences  of  two  circles  are  to  each  other 
as  their  radii,  and  the  areas  of  tivo  circles  are  to  each  other 
as  the  squares  of  their  radii. 

Hyp.  Let  C  and  C  be  the 
Oces,  K  and  R'  the  radii,  and 
S  and  S'  the  areas  of  the  two 
OS. 

To  prove    C  :  C  =  R  :  K', 

and  S:S'=R^R". 

Proof  Inscribe  in  the  Os  two  regular  polygons  of  the 
same  number  of  sides. 

Let  P  and  P'  be  their  perimeters,  and  A  and  A'  their 
areas. 

Then,  because  the  polygons  are  regular,  with  the  same 
number  of  sides, 

P  :  P'  =  R  :  R',  (417) 


and 


A:  A'^R'iR' 


(418) 


Now  let  the  number  of  sides  of  each  polygon  be  increased 

indefinitely,  the  number  remaining  always  equal  in  each. 

Then  P  and  P'  will  approach  C  and  C  as  their  limits,  (431) 

and    A  and  A'  will  approach  S  and  S'  as  their  limits.  (431) 

Since  the  above  ratios  remain  the  same  whatever  be  the 

number  of  sides  in  the  polygons,  so  long  as  the  number  is 

the  same  in  each  (417),  their  limits  are  in  the  same  ratios. 

(425) 


C  :  C  =  R  :  R',  and  S  :  S'  =  R^  R'^ 


434.  CoK.  1.   ^,  = 


and 


S_ 
S' 


4R^ 


4R"' 


Q.E.D. 

(294) 


C   _2R 
C'  ~  2R'^ 

Therefore,  the  circumferences  of  circles  are  to  uich  other 
as  their  diameters,  and  their  areas  are  to  each  other  as  the 
squares  of  their  diameters. 


BOOK  v.- VALUE  OF  7t.  221 

435.  Cor.  2.  Since  similar  arcs  and  sectors  are  like 
parts  of  the  respective  Oces  and  Os  to  which  they  belong 
(432),  therefore  : 

Similar  arcs  are  to  each  other  as  their  radii,  and  similar 
sectors  are  to  each  other  as  the  squares  of  their  radii, 

436.  Cor.  3.  By  alternation,  we  have,  from  (434), 

C  :  2  R  =  C  :  2  R'. 

That  is,  the  ratio  of  one  Oce  to  its  diameter  is  the  same 
as  the  ratio  of  any  other  Oce  to  its  diameter.     Therefore  : 

The  ratio  of  the  circumference  of  a  circle  to  its  diameter 
is  constant. 

This  constant  ratio  is  usually  represented  by  the  Greek 
letter  n,  so  that 


—  =  7r;  .-.  C  =  2;rR. 


437.  ScH.  The  ratio  7r_is  incommensurable,  and  can  V/ 
therefore  be  expressed  only  approximately  in  numbers,  /> 
though  the  letter  n  is  used  to  represent  its  exact  value. 

We  give  here  the  value  of  tt,  its  reciprocal,  and  its 
logarithm: 

It  -  3.14159  26535  89793  23846 

'I 

-  =  0.31830  98861  83790  67153 

n 


log  7t  =  0.49714  98726  94133  85435 


222 


PLANE  GEOMETliY. 


Proposition  9.    Theorem. 

438.  TJie  area  of  a  circle  is  equal  to  half  the  product  of 


A-       ^7 

radius,         U 


its  radius  and  circumference, 
Hyp,  Let  S  be  the  area,  K  the 

and  0  the  Oce  of  the  O. 

To  prove  S  =  J  R  X  C. 

Proof  Circumscribe  about  the  O  auy 

regular  polygon;  let  P  be  its  perimeter, 

and  A  its  area. 


Then 


A  =  J  P  X  11. 


(419) 


Now  let  the  number  of  sides  of  the  polygon  be  increased 
indefinitely. 

Then  the  perimeter  of  the  polygon  will  approach  the  Oce 
of  the  O  as  its  limit,  (431) 

and  the  area  of  the  polygon  will  approach  the  ai-ea  of 
the  O  as  its  limit.  (431) 

But  the  above  equality  remains  true  whatever  be  the 
number  of  sides  of  the  polygon. 

•  • .  in  the  limit  we  have 


S=:iRxC. 

439.  Cor.  1.  By  (436)  C  =  2;rR. 
Substituting  above,  we  obtain 


(425) 

Q.E.D. 


Therefore,  the  area  of  a  circle  is  equal  to  the  square  of 
the  radius  multiplied  by  the  constant  ratio  it. 

440.  Cor.  2.  Let  s  denote  the  area  of  a  sector,  and  c 
the  arc. 


BOOK  V.-THE  MEASURE  OF  TUE  CIRCLE. 


223 


Then,  since  the  sector  is  the  same  part  of  the  O  that 
the  arc  is  of  the  Oce, 

r.s'.^  =  c'.Q,  (294) 


or 


.      _  g  X  S  _  g  X  jR  X  C 

s  =  -JR  X  g. 


(438) 


Therefore,  the  area  of  a  sector  is  equal  to  half  the  product 
of  its  radius  and  arc. 

Proposition  lO.    Theorem. 

441.  The  areas  of  tivo  similar  segments  are  to  each 
other  as  the  squares  of  their  radii. 

Hyp,  Let  S  and  S'  be  the 
areas  of  the  two  similar  sectors 
AOB,  A'O'B',  T  and  T'  the 
areas  of  the  two  similar  As 
AOB,  A'O'B',  R  and  R'  the 
radii. 


To  prove 

S'-T'~R'*' 

Proof,  Since 

Z0=  ZO', 

(432) 

.•.S:S'  =  R^R'% 

(435) 

and 

T:T'  =  R»:R'^  . 

(377) 

ST        R' 
•*'S'~  T'~R"* 

• 

S-T        R'     ^    ^.  .  .     , 
.  g/  _  ry^,  =  j^  (by  division). 

Q.E.D. 

EXERCISES. 

1.  What  is  the  area  of  a  circle  whose  radius  is  40  feet  ?tnf^ 

2.  What  is  the  diameter  of  a  circle  whose  circumference 
is  57  yards? 


224 


PLANE  GEOMETRY. 


Pkoblems  of  Construction. 


Proposition  1  1 .    Problem. 

442.  To  inscribe  a  square  in  a  given  circle. 
Given,  the  O  ABCD,  with  centre  0. 
Required,  to  inscribe  a  square  in  it. 
Cons.  Draw  the  diameters  AC,  BD 

J_  to  each  other. 

Join  AB,  BC,  CD,  DA. 

Then  ABCD  is  the  square  required. 

Proof,  Since  the   Zs  at  0  are  all 
rt/s,  (Cons.) 

.*.  the  sides  AB,  BC,  etc.,  are  equal, 
in  the  same  0  equal  Zs  at  the  centre  intercept  equal  chords  (196), 

and  the  Z  s  BAD,  ADC,  etc.,  are  rt.  Z  s, 
evei'y  £  inscribed  in  a  \Q  is  art.  Z  (240). 

.*.  the  figure  ABCD  is  a  square.  (120) 

Q.E.F. 

443.  Cor.  1.  If  tangents  he  draiun  to  the  circle  at  the 
points  A,  B,  C,  D,  the  figure  so  formed  will  he  a  circum- 
scrihed  square. 

444.  Cor.  2.  To  inscribe,  and  circumscrihe,  regular 
polygons  of  8  sides,  bisect  the  arcs  AB,  BC,  CD,  DA,  and 
proceed  as  before. 

By  repeating  this  process,  regular  inscribed  and  circum- 
scribed 2Jolygons  of  16,  32, ....  ,  and,  in  general,  of  2n 
sides,  may  he  drawn, 

_  EXERCISES. 

7 

^       1.  What   is  the  area  of   a  square  inscribed  in  a  circle 
whose  area  is  48  feet  ? 

2.  What  is  the  area  of  a  regular  hexagon  inscribed  in  a 
circle  whose  area  is  560  square  feet  ? 


BOOK   v.— PROBLEMS  OF  CONSTRUCTION.        22o 


Proposition  12.    Problem. 

445.  To  inscribe  a  regular  hexagon  in  a  given  circle. 
Given,  the  O  ABC,  with  centre  0. 
Required,  to  inscribe  a  regular  hexa- 
gon in  it. 

Cons.  With  any  pt.  A  on  the  ©ce  as   pi 

a  centre,  and  AO  as  a  radius,  describe 

an  arc  cutting  the  ©ce  in  B. 
Join  AB,  BO,  OA.  ^ 

Then  AB  is  a  side  of  the  hexagon  required. 
Proof.  Since  the  A  OAB  is  equilateral,  (Cons.) 

.-.  it  is  equiangular.  (113) 

.-.   Z  AOB  is  4  of  2  rt.  Z  s,  or  |  of  4  rt.  I  s.  (103) 

.*.  the  arc  AB  is  \  of  the  Oce. 

.*.  the  chord  AB  is  a  side  of  the  regular  inscribed  hexa- 
gon. (400) 

.-,  the  figure  ABCDEF,  completed  by  drawing  the  chords 
BC,  CD,  DE,  EF,  FA,  each  equal  to  the  radius  OA,  is  the 
regular  inscribed  hexagon  required.  Q.E.F. 

446.  CoK.  1.  If  the  alternate  vertices  of  the  regular  hex- 
agon he  joined,  we  obtain  the  inscribed  equilateral  triangle 
ACE. 

447.  Cor.  2.  If  tangents  be  drawn  to  the  circle  at  the 
poi7its  A,  B,  C,  D,  E,  F,  the  figure  so  formed  will  be  a  reg- 
ular circumscribed  hexagon. 

448.  Cor.  3.  To  inscribe,  and  circumscribe^  regular 
polygons  of  12  sides,  bisect  the  arcs  AB,  BC,  CD,  etc.,  and 
proceed  as  before. 

By  repeating  this  process,  regular  inscribed  and  circum^ 
scribed  polygons  of  24,  48,  etc.,  sides  may  be  drawn. 


m 


PLANE  GEOMETRY. 


Proposition  13.    Problem. 

449.  To  inscribe  a  regular  decagon  in  a  given  circle. 

Given,  the  O  ABCE. 

Reqtiired,  to  inscribe  in  it  a  regular  dec- 
agon. 

Cons.  Draw  any  radius  OA,  and  divide 
it  in  extreme  and  mean  ratio  at  D,  so 
that 

OA  :  OD  =  OD  :  AD.     (352) 
With  centre  A  and  OD  as  a  radius,  describe  an  arc  cut- 
ting the  Oce  at  B,  and  join  AB. 

Then  AB  is  a  side  of  the  required  inscribed  decagon. 
Proof.  Join  BO,  BD. 


Since 
and 


OA 


(Cons.) 
(Cons.) 


OD  =  OD  :  AD, 

OD  =  BA, 

.  • .  OA  :  BA  =  BA :  AD. 

.  • .  the  A  s  OAB  and  BAD  are  similar, 
having  Z  A  common  and  the  including  sides  proportional  (314). 

.-.   ZABD=  ZAOB. 
Because 
OA  =  OB,  and  BA  =  OD, 

.  • .  OB  :  BA  =r  OD  :  AD, 

.-.   ZABD=  zDBO. 
.-.   ZAB0  =  2  ZABD  =  2  zAOB. 
.-.  also  ZBAO  =  ZAB0  =  2  ZAOB, 

being  opp.  the  equal  sides  of  an  isosceles  A  (111). 
Z  ABO  +  Z  BAO  H-  Z  AOB  =  5  Z  AOB  =  2  rt.  Z  s. 
Z  AOB  =  I  of  2  rt.  Z  s,  or  ^o  of  4  rt.   Z  s. 
the  arc  AB  is  j\  of  the  Oce. 

the  chord  AB  is  a  side  of  the  regular  inscribed  deca- 

(400) 
the  figure  ABCE  .  ..  .  ,  formed  by  applying  AB  ten 
times  to  the  Oce,  is  the  regular  inscribed  decagon  required. 

Q.E.F. 


(307) 

(Radii)  and  (Cons.) 

(303) 


gon 


BOOK  v.— PROBLEMS  OF  CONSTRUCTION.        227 

450.  Cor.  1.  If  the  alternate  vertices  of  the  required 
decagon  be  joi7ied,  a  regular  pentagon  is  inscribed. 

451.  Cor.  2.  If  tangents  be  drawn  at  the  points  atiuhich 
the  circumference  is  divided,  the  figure  so  formed  ivill  be  a 
regular  circumscribed  decagon. 

452.  Cor.  3.  To  inscribe,  and  circumscribe,  regular 
polygons  of  20  sides,  bisect  the  arcs  AB,  BC,  etc.,  and  pro- 
ceed as  before. 

By  repeating  this  process,  regular  i?iscribed  and  circum- 
scribed polygons  0/40,  80,  etc.,  sides  may  be  draion. 

EXERCISES. 

1.  The  side  of  an  inscribed  square  is  equal  to  the  radius 
of  the  circle  multiplied  by  \^2. 

2.  The  side  of  an  inscribed  equilateral  triangle  is  equal 
to  the  radius  multiplied  by  V^. 

3.  The  apothem  of  an  inscribed  square  is  equal  to  half 

the  radius  multiplied  by  V^, 

4.  The  apothem  of  an  inscribed  equilateral  triangle  is 
equal  to  half  the  radius. 

5.  The  apothem  of  a  regular  inscribed  hexagon  is  equal 
to  half  the  radius  multiplied  by   V^. 

6.  The  area  of  a  circumscribed  square  is  double  the  area 
of  the  inscribed  square. 

7.  Kequired  the  area  of  an  equilateral  triangle  inscribed 
in  a  circle  whose  radius  is  4. 

8.  Kequired  the  area  of  a  square  inscribed  in  a  circle 
whose  radius  is  5. 

9.  Kequired  the  area  of  a  regular  hexagon  inscribed  in  a 
circle  whose  radius  is  8. 

10.  Kequired  the  area  of  a  circle  whose  circumference  is 
100. 

11.  Kequired  the  area  of  a  circle  inscribed  in  a  square 
whose  area  is  36, 


228  PLANE  GEOMETRY. 

Proposition.  1 4.    Problem. 

453.  To  inscrihe  a  regular  pentadecagon  in  a  given  cir- 
cle. 

Given,  the  O  ABC.  /- ->^ 

Required,  to  inscribe  in  it  a  regular      /  ^ 

polygon  of  15  sides.  /  Ae 

Cons.     Lay  off  HB  equal  to  a  side     I  L 

of  a  regular  inscribed  hexagon.     (445)     V  / 

Lay  off  HA  equal  to  a  side  of  a  regular       \^ -3^^^ 

inscribed  decagon.  (449)         *^^""^^^^A**^ 

Join  AB. 

Then  AB  is  a  side  of  the  required  inscribed  pentadeca- 
gon. 

Proof,     Arc  AB  =  arc  HB  —  arc  HA 

=  -J-  of  Oce  —  ^V  of  Oce  =  ^^  of  Oce. 

.  • .  the  chord  AB  is  a  side  of  the  regular  inscribed  penta- 
decagon. (400) 

.'.the  figure  ABCD  . .  . ,  formed  by  applying  AB  15 
times  to  the  Oce,  is  the  regular  inscribed  pentadecagon 
required.  q.e.f. 

454.  CoK.  1.  If  tangents  be  dratun  at  the  points  at 
which  the  circumference  is  divided,  a  regular  circum- 
scrihed  pentadecagon  is  obtained. 

455.  Cor.  2.  To  inscribe,  and  circumscribe,  regular 
polygons  of  30  sides,  bisect  the  arcs  AB,  BC,  etc.,  and 
proceed  as  before. 

By  repeating  this  process,  regular  inscribed  and  circum- 
scribed polygons  q/*60,  120,  etc.,  sides  may  be  draiun. 

456.  ScH.  These  are  the  only  polygons  that  could  be 
constructed  by  the  ancient  geometers,  by  the  use  of  the 
rule  and  compass.  About  the  beginning  of  the  present 
century,  Gauss,  the  great  German  mathematician,  proved 
that  whenever  2"  +  1  is  a  prime  number,  and  n  an  integer, 
a  regular  polygon  of  that  number  of  sides  could  be  in- 
scribed in  the  circle,  by  the  rule  and  compass. 

Therefore,  it  is  possible  to  inscribe  regular  polygons  of. 


BOOK  v.— PROBLEMS  OF  COMrVTATION. 


229 


17  sides,  and  of  257  sides  ;  but  the  construction  of  the  lat- 
ter polygon  is  so  lengthy,  it  is  not  likely  that  it  has  ever 
been  attempted. 


Proposition  1  5.    Problem. 

45  7.  Give7i  the  radius  and  tlie  side  of  a  regular  inscribed 
polygon,  to  compute  the  side  of  a  similar  circumscribed 
poly g  071. 

Given,  AB  a  side  of  the  regular  in-   ^^ 
scribed    polygon,     and    OF  =  K,    the 
radius  of  the  O  ABH. 

Required,  to  compute  CD,  a  side  of 
the  similar  circumscribed  polygon. 

Cons.    Join  00,  OD. 

They  will  cut  the  Oce  in  A  and  B. 

The  similar  as  OOF,  OAE  give 


OF 
AE 


OF 
OE* 


4=^=:^.    .-.CD^ 


RX  AB 
OE     • 


In  the  rt.  A  OAE,     OE  =  Voa'  — AE'  ' 


.  OE 


=  y^ 


/AB 


CD  = 


2Rx  AB 


458.  SCH. 


(307) 

(328) 


Q.E.F. 


V  4R^  -  AB^ 
When  R  =  1,  this  becomes 

If  we  know  the  sides  of  the  regular  polygons  of  n,  2n, 
4:n,  ....  sides,  inscribed  in  the  circle  of  radius  1,  we  have, 
by  this  formula,  the  sides,  and  therefore  the  perimeters, 
of  the  regular  circumscribed  similar  polygons. 


230 


PLANE  GEOMETRY. 


Proposition  16.    Problem. 

459.  Given  the  radius  and  the  side  of  a  regular  in- 
scribed  polygon,  to  compute  the  side  of  the  regular  in- 
scribed poly  go7i  of  double  the  number  of  sides. 

Given,  AB,  a  side  of  the  regular  in-  ^ 

scribed  polygon,  and  00  =  R,  the  radius    A/ 
of  the  O  ABD. 


Required,  to  compute  AO,  a  side  of  the 
regular  inscribed  polygon  of  double  the 
number  of  sides. 

Cons,     0  is  the  mid.  pt.  of  the  arc  AB. 

Produce  00  to  D,  and  join  OA. 

OD  is  _L  to  AB  at  its  mid.  pt. 

AO  is  a  mean  proportional  between  OD  and  OE 

.-.  AO"^  =  OD  X  OE  =  R  (2R  -  20E). 

But  OE  =  i  \/4R»  -^^. 

.-.  AO  =  |/r  (2R  -  V'iw  -  A&) 
460.  ScH.     When  R  =  1,  this  gives 


(403) 


(204) 
(325) 


(457) 

Q.E.F. 


AQ  =  \/'2-  V'4-AB' 


By  repeated  applications  of  this  formula,  we  may  com- 
pute successively  the  sides,  and  therefore  the  perimeters, 
of  the  regular  inscribed  polygons  of  2n,  4:n,  871,  16n, .... 
sides,  n  being  the  number  of  sides  of  the  first  polygon. 

EXERCISES. 

1.  If  the  radius  of  a  circle  is  4,  find  its  circumference 
and  area. 

2.  If  the  circumference  of  a  circle  is  28,  find  its  diameter 
and  area. 


BOOK   V— METHOD   OF  rEUIMETERS. 


231 


Proposition   1 7.    Problem. 

461.  To  conqmte  the  ratio  of  the  circumference  of  a  cir- 
cle to  its  diameter. 

Given,  the  Oce    C,  iind  the  radius  K. 

Required,  to  find  the  number  tt. 

Cons,  C  =  2;rk  (436) 

Let      R  —  1,  then  n  =  |C. 

That  is,  the  number  rr  =  semi  Oce  of  radius  1. 

Therefore,  the  semi-perimeter  of  each  polygon  inscribed 
in  this  O  ce  is  an  approximate  vakie  of  tt  :  and  as  the  num- 
ber of  sides  of  the  polygons  increases  indefinitely,  the 
lengths  of  the  perimeters  approach  to  that  of  the  Oce  as 
the  limit.  (430) 

Hence,  by  the  process  of  (4G0),  we  may  obtain  a  succes- 
sion of  nearer  and  nearer  approximations  to  the  length  of 
the  semi  Oce. 

It  is  convenient  to  begin  the  computation  with  the  in- 
scribed hexagon. 

.  •.  making  AB  =  1,  we  have  from  (460),  the  following: 


Number  of 
Sides. 

Semi-perimeters. 

Number  of 
sides. 

Semi-perimeter 

6 

3.00000000 

96 

3.14103198 

12 

3.10582854 

192 

3.14145255 

24 

3.13262861 

384 

3.14155772 

48 

3.13935026 

768 

3. 1415847  L 

The  last  two  results  show  that  the  first  four  decimals  do 
not  change  as  the  number  of  sides  is  increased. 

Hence  the  approximate  value  of  tt  is  3.1415,  correct  to 
the  fourth  decimal  place.  q.e.f. 

In  practice  we  generally  take  tt  =  3.14159. 


232  PLANE  GEOMETRY. 

462.  ScH.  The  above  is  called  the  method  of  perimeters, 
For  the  meiliod  of  isoperimeters  seeRouche  et  €omberousse. 
Edition  of  1883,' p.  194. 

Note.— The  number  n  is  of  such  fundamental  importance  in  geometry  that 
mathematicians  have  sought  for  its  value  in  a  great  variety  of  ways,  all  of 
which  agree  in  the  conclusion  that  it  cannot  be  expressed^  exactly  in  decimals, 
but  only  approximately.  Aichimedes  (born  287  B.C.)  was  the  first  to  assign  an 
approximate  value  of  ir.  He  proved  tliat  it  is  included  between  the  numbei*s 
Z\  and  m,  or,  in  decimals,  between  3.1428  and  3.1408;  he  therefore  assigned  its 
value  correctly  within  a  unit  of  the  third  decimal  place.  The  number  3f,  or  V 
is  often  used  in  rough  computations. 

Adrian  Metius,  a  Dutch  geometer  of  the  16th  century,  has  given  us  the  much 
more  accurate  value  of  f  f  §,  which  is  correct  to  within  a  half-millionth,  and 
which  is  remarkable  for  the  manner  in  which  it  is  formed  by  the  first  three  odd 
numbers  1,  3,  5,  each  written  twice. 

More  recently,  the  value  has  been  found  to  a  great  number  of  decimals  by  the 
aid  of  series.  Dase  and  Clausen,  German  computers,  carried  the  calculation  to 
200  decimal  places,  independently  of  each  other,  and  their  results  agreed  to  the 
last  figure. 

The  first  20  figures  of  their  result  are  as  follows  : 

3.14159  26535  89793  23846.  (437) 

This  result  is  far  beyond  all  the  wants  of  mathematics. 
Ten  decimals  are  sufficient  to  give  the  circumference  of  tlie 
earth  to  the  fraction  of  an  inch,  and  thirty  decimals  would 
give  the  circumference  of  the  whole  visible  universe  to  a 
quantity  imperceptible  with  the  most  powerful  microscope.* 

EXERCISES. 

1.  The  area  of  the  regular  inscribed  hexagon  is  equal  to 
twice  the  area  of  the  inscribed  equilateral  triangle. 

2.  The  square  of  a  side  of  the  inscribed  equilateral  trian- 
gle is  three  times  the  square  of  a  side  of  the  regular  inscribed 
hexagon. 

3.  The  area  of  a  regular  inscribed  hexagon  is  half  the  area 
of  the  circumscribed  equilateral  triangle. 

4.  If  tlie  diameter  of  a  circle  be  produced  to  C  until  the 
produced  part  is  equal  to  the  radius,  the  two  tangents  from 
C  and  their  chord  of  contact  form  an  equilateral  triangle. 

♦  Newcomb's  Geom.,  p.  235. 


BOOK  V.-EXERCISE8.  233 

5.  Divide  an  angle  of  an  equilateral  triangle  into  five 
equal  parts. 

Describe  a  O  about  tbe  a  ,  then  use  (453). 

6.  The  square  inscribed  in  a  semicircle  is  equal  to  two- 
fifths  the  square  inscribed  in  the  whole  circle. 

7.  The  area  of  a  given  circle  is  314.16;  if  this  circle  be 
circumscribed  by  a  square,  find  the  area  of  the  pai*t  between 
the  circumference  and  the  perimeter  of  the  square. 

8.  The  area  of  a  circle  is  40  feet ;  find  the  side  of  the 
inscribed  square. 

9.  Find  tlie  angle  subtended  at  the  centre  of  a  circl^  by 
an  arc  6  feet  long,  if  the  radius  is  8  feet  long. 

10.  Find  the  length  of  the  arc  subtended  by  one  side  of 
a  regular  octagon  inscribed  in  a  circle  whose  radius  is  20 
feet. 

11.  Find  the  area  of  a  circular  sector  whose  arc  contains 
18",  the  radius  of  the  circle  being  4  feet. 

12.  Find  the  area  of  a  circular  sector,  the  chord  of  half 
the  arc  being  20  inches,  and  the  radius  45  inches. 

13.  The  radius  of  a  circle  is  5  feet  :  find  the  area  of  a 
circle  7  times  as  large. 

14.  The  radius  of  a  circle  is  7  feet :  find  the  radius  of  a 
circle  16  times  as  large. 

15.  Find  the  height  of  an  arc,  the  chord  of  half  the  arc 
being  10  feet,  and  the  radius  16  feet. 

16.  Find  the  area  of  a  segment  whose  height  is  4  inches, 
and  chord  30  inches. 

17.  Find  the  area  of  a  segment  whose  height  is  16  inches, 
the  radius  of  the  circle  being  20  inches. 

18.  Find  the  area  of  a  segment  whose  arc  is  100°,  the  ra- 
dius being  12  feet. 

19.  Find  the  area  of  a  sector,  the  radius  of  the  circle  being 
24  feet,  and  the  angle  at  the  centre  30°. 

20.  A  circular  park  500  feet  in  diameter  has  a  carriage- 
way around  it  25  feet  wide  :  find  the  area  of  the  carriage- 
way. 


234  PLANE  GEOMETRY. 


Maxima  and  Minima. 

46  3 .  Def.  a  maxim  iwi  quantity  is  the  greatest  quantity 
of  the  same  kind  ;  and  a  minimum  quantity  is  the  least 
quantity  of  the  same  kind. 

Thus,  the  diameter  of  a  circle  is  a  maximum  among  all 
inscribed  straight  lines  ;  and  a  perpendicular  is  a  minimum 
among  all  the  straight  lines  drawn  from  a  given  point  to  a 
given  straight  line. 

464.  Isojjerimetric  figures  are  those  which  have  equal 
perimeters. 

We  give  here  a  few  simple  but  important  propositions 
bearing  on  this  part  of  Geometry. 

Proposition    1 8.    Theorem. 

465.  Of  all  triangles  formed  with  two  given  sides,  that 
in  which  these  sides  are  perpendicular  to  each  other  is  the 
maximum. 

Hyp.  Let  ABC,  A'BC  be  two    As 

having  the  sides  AB,  BC,  respectively 

equal  to  A'B,  BO;  and  let  /ABC  be 

a  rt.  Z. 

To  prove  A    ABC  >  A  A'BC.     

Proof  Draw  A'D  _L  to  BC.  B         D  c 

Since  the  _L  is  the  shortest  distance  from  a  pt.  to  a  line,  (58) 

.  • .  A'B  >  A'D. 
But  AB  =  A'B.  (Hyp.) 

.-.  AB>  A'D. 

But  AB  and  A'D  are  the  altitudes  respectively  of  the  A  s 
ABC,  A'BC. 

.  • .   A  ABC  >   A  A'BC.  (369) 

Q.E.D. 


c^"' 

^,x'^ 

! 

iG 

V 

sj._ 

/f 

1 

BOOK  v.— MAXIMA  AND  MINIMA  235 


Proposition  1 9.    Theorem. 

466.  Of  all  triangles  liaving  equal  perimeters  and  the 
same  base,  the  isosceles  triangle  is  the  maximum. 

Hyp.  Let  the  As  ABC,  ABD 
have  equal  perimeters  and  the  same 
base  AB,  and  let  the  A  ABC  be 
isosceles. 

To  2)rove  A  ABC  >  A  ABD. 

Proof,  Produce  AC  to  H,  so  that 
CH  =  CB  =  CA,  and  join  HB. 

Then  ZABHisart. /, 

being  inscribed  in  ilie  |0  wliose  cent,  is  Cand  radius  is  CB  (240). 

Produce  HB,  take  DL  =  DB,  join  AL,  and  draw  CG, 
DM  II ,  and  CE,  DP  J_,  to  AB. 

Then  AD  +  DL  =  AD  +  DB  =  AC  +  CB  =  AH. 

But  AD  +  DL  >  AL, .-.  AH>  AL. 

.•.BH>BL,  (65) 

and  .•.|BH>iBL. 

But    ^  BH  =  BG  =  CE,  and  |BL  =  BM  =  DP.       (61) 

.-.  CE>DF, 

and  .•.aABC>  aABD.  (369) 

Q.E.D. 

467.  CoK.  Of  all  the  triatigles  of  the  same  2^crimeter, 
that  which  is  equilateral  is  the  maximum, 

Por,  the    maximum   triangle  having  a  given  perimeter 
must  be  isosceles  whichever  side  is  taken  as  the  base. 


236  PLANE  GEOMETRY. 


Proposition  20.    Theorem. 

468.  Of  all  isoperinietric  polygons  having  the  same 
number  of  sides ^  the  maximiun  2)olygon  is  equilateral. 

Hyp.  Let  ABODE  be  the  maximum 
polygon  of  given  perimeter  and  given  ^^<^^^^^'^^"- 

number  of  sides.  A^^^^™ -T^i^C 

To  prove  it  is  equilateral.  \  / 

Proof  Join  AC.  \  / 

If  possible,  let  the  sides  AB,  BC  of         A Z 

the  A  ABC  be  unequal,  and  let  AB'C 

be  an  isosceles  A  having  the  same  base  AC,  and 

AB'-f  B'C  =  AB  +  BC. 

Then,  aAB'C>aABC.  (466) 

But        the  area  ACDE  remains  unchanged. 

.  • .  area  AB'CDE  >  area  ABCDE.         (Ax.  4) 

But  area  ABCDE  is  the  maximum.  (Hyp.) 

.  • .  area  AB'CDE  <  area  ABCDE, 

and  .*.  AB  and  BC  cannot  be  unequal. 

.-.  AB  =  BC. 

In  the  same  way  it  may  be  proved  that 

BC  =  CD  =  DE  =  etc.  Q.E.D. 

EXERCISE. 

If  the  diagonals  of  a  quadrilateral  are  given  in  magnitude, 
the  area  of  the  quadrilateral  is  a  maximum  when  the  diago- 
nals are  at  right  angles  to  each  other. 


BOOK  v.— MAXIMA  AND  MINIMA.  237 


Proposition  2 1 .    Theorem. 

469.  Of  all  2)olygons  formed  of  sides  all  given  hut  one, 
the  maximum  can  be  inscribed  in  a  semicircle  with  the  un- 
determined side  for  its  diameter.* 

Hyp.  Let  ABCDEF  bo  the 
maximum  polygon  formed  of  the 
given  sides  AB,  BC,  CD,  DE,  EF. 

To  j^trove  ABCDEF  can  be  in- 
scribed in  a  semicircle. 

« 
Proof  Join  any  vertex,  as  D,  with  the  extremities  A  and 

F  of  the  side  not  given. 

Then,  the  A  ADF  must  be  the  max.  of  all  A  s  formed 
witli  the  given  sides  AD  and  FD;  for,  if  it  is  not,  by  in- 
creasing or  decreasing  the  /  ADF,  keepmg  tlie  sides  AD 
and  FD  unchanged  in  length,  the  A  ADF  may  be  increased, 
wliile  the  rest  of  the  polygon,  ABCD,  DEF,  remains  un- 
changed; so  that  the  whole  polygon  ABCDEF  is  increased. 

But  this  is  contrary  to  the  hypothesis  that  the  given 
polygon  is  a  maximum. 

.  • .  the  A  ADF  must  be  the  max.  of  As  formed  with  the 
given  sides  AD  and  FD. 

.  • .  the  Z  ADF  is  a  rt.  Z .  (4G5) 

.  • .  D  IS  on  the  semiOce  whose  diameter  is  AF.        (240) 

.*.  any  vertex  is  on  the  semiOce  whose  diameter  is  AF. 

Q.E.D. 

*  Peirce's  Geometry,  p.  92 


238 


PLANE  GEOMETRY. 


Proposition  22.    Theorem. 

470.   Of  all  polygons  formed  loith  the  same  given  sides, 
that  which  can  be  inscribed  in  a  circle  is  the  maximum. 

Hyp,  Let  ABODE  be  a 
polygon  inscribed  in  a  O, 
and  A'B'C'D'E'  any  other 
polygon  with  the  same  sides 
as  the  first,  but  which  can- 
not be  inscribed  in  a  O. 

To  prove 
ABODE  >A'B'C'D'E'. 

Proof  Draw  the  diameter  DH. 

Join  AH,  HB. 

Upon  A'B'  (=  AB)  cons.  aA'B'H'  =  aABH,  and  join 
D'ir. 

Then,  since  the  polygon  HAED  is  inscribed  in  \0  with 
diameter  HD, 


and 

Adding, 
Subtracting, 


.-.  HAED  >  H'A'E'D', 
HBCD  >  H'B'C'D'. 
AHBCDE  >  A'H'B'O'D'E'. 

AABH=r  aA'B'H'. 
•.  ABODE  >  A'B'O'D'E'. 


(469) 
(469) 


(Ax.  5) 

Q.E.D. 

471.  Cor.  The  maximum  of  all  isoperimetric  polygons 
of  the  same  number  of  sides  is  regular. 

For,  it  is  equilateral,  (468) 

and  it  can  be  inscribed  in  a  O,  (470) 

.  • .  the  polygon  is  regular.  (400) 

EXERCISE. 

Find  a  point  in  a  given  straight  line  such  that  the  tan- 
gents drawn  from  it  to  a  given  circle  contain  the  greatest 
angle  possible. 


BOOK  v.— MAXIMA  AND  MINIMA.  239 


Proposition  23.    Theorem. 

472.   Of  regular  polygons  ivith  a  given  perimeter,  that 
tvhicJi  has  the  greatest  number  of  sides  has  the  greatest  area. 

Hyp.  Let  P  be  a  regular  poly- 
gon of  three  sides,  and  Q  a  regular 
polygon  of  four  sides,  with  the 
same  given  perimeter. 

To  2)rove  Q  >  P. 


Q 


Proof.  In  any  side  AB  of  P  take  any  pt.  C. 

The  polygon  P  may  be  regarded  as  an  irregular  polygon 
of  four  sides,  in  which  the  sides  AC,  CB  make  with  each 
other  a  st.  Z . 

Then,  since  the  polygons  P  and  Q  are  isoperimetric,  and 
have  the  same  number  of  sides,  (HyP-) 

.  • .  the  irregular  polygon  P  <  the  regular  polygon  Q.  (471) 

In  the  same  way  it  may  be  shown  thut  Q  <  the  i-egular 
isoperimetric  polygon  of  five  sides,  and  so  on.  q.e.d. 

473.  Cor.  Tlie  circle  has  a  greater  area  than  any  poly- 
gon of  the  saine  p)&rimeter, 

EXERCISES. 

1.  Of  all  triangles  of  given  base  and  /'' 
area,  the  isosceles  is  that  which  has  the  /  / 
greatest  vertical' angle.  1  // 

2.  The    shortest   chord   which   can  be  \ — 

drawn   through   a   given   point   within   a  ^^^ --'' 

circle  is  the '  perpendicular  to  the  diameter  which  passes 
through  tliat  point. 


240 


PLANE  GEOMETRY. 


Proposition  24.    Theorem. 

474.   Of  regular  polygons  having  ihe  same  given  area, 
the  greater  the  number  of  sides  the  less  ivill  he  theiierimeivw 

Hyp.  Let  P  and  Q  be  regular 
polygons  having  the  same  area, 
and  let  Q  have  the  greater  number 
of  sides. 

To  prove  the  perimeter  of  P  > 
that  of  Q. 

Proof.  Let  R  be  a  regular  poly- 
gon having  the  same  perimeter  as 
Q  and  the  same  number  of  sides  as  P. 

Then,  since  Q  has  the  same  perimeter  as  R  and  a  greater 

number  of  sides,  (Cons.) 

.•.Q>R.  (4;'.>) 

But  Q=P,  (Hyp.) 

.-.  P>  R. 

.*.  the  perimeter  of  P  >  the  perimeter  of  R.     (370) 

.  •.  the  perimeter  of  P  >  the  perimeter  of  Q.    q.e.d. 

475.  Cor.    The  circumference  of  a  circle  is  less  than  the 

perimeter  of  any  polygon  of  the  same  area. 


Proposition  25.    TFieorem. 

476.  Given  two  intersecting  straight  lines  AB,  AC,  and 
a  point  P  between  them  ;  then  of  all  straight  lines  which 
pass  through  P  and  are  terminated  hy  AB,  AC,  that  ivhich 
is  bisected  at  P  cuts  off  the  triangle  of  minimum  area. 

Hyp.  Let  EF  be  the  st.  line,  termi- 
nated by  AB,  AC,  which  is  bisected  f> 
at  P. 

To  prove  A  AEF  is  a  minimum. 

Proof    Let  HK  be    any  other  st.    a" 
line  passing  through  P. 


» 


BOOK  v.— MAXIMA  AND  MINIMA.  241 

Through  E  draw  ED  ||  to  AC. 

Then  aHPF  =  aDPE.  (105) 

But  aDPE  <  aKPE. 

.-.  aHPF  <  aKPE. 
Add  area  AEPII  to  each. 

/.  aAEF  <  aAKH.  (Ax.  4) 

In  the  same  way  it  may  be  shown  that  A  AEF  <  any 
otlier  A  formed  by  a  st.  line  through  P. 

.*.  A  AEF  is  a  minimum.  q.e.d. 


Proposition  26.    Problem. 

477.  To  find  at  what  point  in  a  given  straight  line  the 
angle  subtended  hy  the  line  joining  two  given  points,  lohich 
are  on  the  same  side  of  the  given  straight  line,  is  a  maximum. 

Given,  the  st.  line  CD,  and  the  pts. 
A,  B,  on  the  same  side  of  CD. 

Required,  to  find  at  what  pt.  in  CD 
the  Z  subtended  by  the  st.  line  AB  is 
a  maximum. 


Cons.  Describe  a  O  to  pass  through  " '' 

A,  B,  and  to  touch  the  st.  line  CD.  (339)  and  Ex.  40  in  (354) 

Let  P  be  the  pt.  of  contact. 

Then  the  Z  APB  is  the  required  max.  Z . 

Proof.  Take  any  other  pt.  in  CD  as  Q,  and  join  AQ,  BQ. 

Then,  Z  AQB  <  Z  APB.  (246) 

.  • .  Z  APB  >  any  other  Z  subtended  by  AB  at  a  pt.  in  CD. 

Q.E.F. 


242  PLANE  GEOMETRY. 


Proposition  27.    Problem. 

478.  In  a  straight  line  of  indefinite  length  find  a  point 
such  that  the  sum  of  its  distances  from  two  given  point f<, 
on  the  same  side  of  the  given  line,  shall  be  a  mininmm. 

Given,  the  st.  line  CD,  and 
tlie  pts.  A,  B,  on  the  same  side  f"''^; 

of  CD. 

Required,  to  find  a  pt.  P  in     C- 
CD,  so  that  the  sum  of  AP,  PB 
is  a  minimum. 

Cons.     Draw    AF  J.  to  CD; 


P        ,- 


and  produce  AF  to  E,  making  FE  =  AF. 

Join  EB,  cutting  CD  at  P. 

Join  AP,  PB. 

Then,  P  is  the  required  pt.,  and  of  all  lines  drawn  from 
A  and  B  to  a  pt.  in  CD,  the  sum  of  AP,  PB  is  a  minimum. 

Proof.    Let  Q  be  any  other  pt.  in  CD. 

Join  AQ,  BQ,  EQ. 

Then,  rt.  A  AFP  =  rt.  A  EFP.  (104) 

.-.  AP  =  EP. 

Similarly,  AQ  =  EQ. 

But  EB  <  EQ  +  QB.  (96) 

.-.  AP  +  PB  <  AQ  +  QB. 

.  • .  the  sum  of  AP  and  PB  is  a  minimum.      q.e.f. 

479.  CoE.  TJte  sum  of  AP  and  PB  is  a  minimum,  when 
these  lines  are  equally  inclined  to  CD; 

for  Z  APC  =  Z  EPC  =  Z  BPD. 

Note.— In  order  that  a  ray  of  light  from  A  may  be  reflected  to  a  point  B,  it 
must  fall  upon  a  mirror  CD  at  a  point  P  where  ZAPC  =  /BPD  ;  i.e.,  by  (478) 
the  ray  pursues  the  shortest  path  between  A  and  B  and  touching  CD. 


BOOK  v.— EXERCISES.     THEOREMS.  243 


exercises. 
Theorems. 


1.  If  AB  be  a  side  of  an  equilateral  triangle  inscribed  in 
a  circle,  and  AD  a  side  of  the  inscribed  square,  prove  that 
three  times  the  square  on  AD  is  equal  to  twice  the  square 
on  AB. 

2.  Show  that  the  sum  of  the  perpendiculars  from  any 
point  inside  a  regular  hexagon  to  the  six  sides  is  equal  to 
tliree  times  the  diameter  of  the  inscribed  circle. 

3.  The  area  of  the  regular  inscribed  hexagon  is  twice  the 
area  of  the  inscribed  equilateral  triangle. 

4.  The  area  of  the  regular  inscribed  hexagon  is  three- 
fourths  of  that  of  the  regular  circumscribed  hexagon. 

5.  The  area  of  the  regular  inscribed  hexagon  is  a  mean 
proportioual  between  the  areas  of  the  inscribed  and  circum- 
scribed equilateral  triangles. 

G.  If  the  perpendicular  from  A  to  the  side  BC  of  the 
equilateral  triangle  ABC  meet  BC  in  D,  and  the  inscribed 
circle  in  G;  prove  that  GD  =  2 AG. 

See  figure  of  (269). 

7.  If  three  circles  touch  each  other  externally,  and  a  tri- 
angle be  formed  by  joining  their  centres,  and  another  tri- 
angle by  joining  their  points  of  contact,  the  inscribed  circle 
of  the  former  triangle  will  be  the  circumscribed  circle  of 
the  latter. 

8.  If  ABCD  be  a  square  described  about  a  given  circle, 
and  P  any  point  on  the  circumference  of  the  circle;  prove 
that  the  sum  of  the  squares  on  PA,  PB,  PC,  PD,  is  three 
times  the  square  on  the  diameter  of  the  circle. 

Use  (3:33). 

9.  ABC  is  a  triangle  having  each  of  the  angles  B,  C 
double  the  angle  A;  the  bisectors  of  the  angles  B,  C  meet 
AC  and  the  circle  circumscribing  the  triangle  ABC  respec- 
tively in  D,  E :  prove  that  ADBE  is  a  rhombus. 

10.  ABCDE  is  a  regular  pentagon  inscribed  in  a  circle, 
P  is  the  middle  point  of  the  arc  AB:  show  that  the  dif- 
ference of  the  straight  lines  AP,  CP  is  equal  to  the  radius 
of  the  circle. 


244  PLANE  GEOMETRY. 

11.  In  the  last  exercise  prove  that  the  sum  of  the  squares 
on  PC  and  BC  is  equal  to  four  times  the  square  on  the 
radius. 

12.  ABCDE  is  a  regular  pentagon,  BE  is  drawn  cutting 
AC^  AD  in  F,  G  respectively :  show  tliat  the  sum  of  AB 
and  AE  is  equal  to  the  sum  of  BE  and  FG. 

Describe  a  ®  about  the  pentagon,  etc. 

13.  In  a  circle  a  regular  pentagon  and  a  regular  decagon 
are  inscribed;  the  middle  points  of  the  adjacent  sides  of 
the  pentagon  are  joined:  prove  that  the  sides  of  the  penta- 
gon so  formed  are  equal  to  the  radius  of  the  circle  inscribed 
in  the  decagon. 

14.  Every  equilateral  polygon  inscribed  in  a  circle  is 
equiangular. 

15.  Every  equilateral  polygon  circumscribed  about  a  cir- 
cle is  equiangular,  if  the  number  of  sides  be  odd. 

16.  Every  equiangular  polygon  inscribed  in  a  circle  is 
equilateral,  if  the  number  of  sides  be  odd. 

17.  Every  equiangular  polygon  circumscribed  about  a 
circle  is  equilateral. 

18.  The  figure  formed  by  the  five  diagonals  of  a  regular 
pentagon  is  another  regular  pentagon. 

10.  If  the  alternate  sides  of  a  regular  pentagon  be  pro- 
duced to  meet,  the  five  points  of  meeting  form  another  reg- 
ular pentagon. 

Let  a  denote  a  side  of  a  regular  polygon  inscribed  in  a 
circle  whose  radius  is  R;  then; 

20.  In  a  regular  decagon, 


a  =  f(4/5-l). 

Jse  (449\ 

31.  In  a  regular  pentagon, 

T?      / 

' 

a^|VlO-2V5. 

(459  reciprocally) 

22.  In  a  regular  octagon. 

a  =  R  V2  -  V2. 

(459) 

BOOK  V.    NUMElilCAL  EXERCISES.  245 

23.  In  a  regular  dodecagon, 

a  =  11/2-4/3.  (459) 

24.  In  a  regular  pentad ecagon, 

«  =  ?(  /lO  +  2  V5  +  V^-  VWj.  (453) 

25.  The  side  of  the  regular  inscribed  pentagon  is  equal 
to  the  hypoten^ise  of  a  right  triangle  whose  sides  are  the 
radius  and  the  side  of  the  regular  inscribed  decagon. 

Numerical  Exercises. 

26.  The  diameter  of  a  circle  is  5  feet;  find  the  side  of 
the  inscribed  square.  Aris.  3.535  ft. 

27.  The  apothem  of  a  regular  hexagon  is  2;  find  the  area 
of  the  circumscribing  circle.  Ans,  b\  7t, 

28.  There  are  two  gardens;  one  is  a  square,  and  the 
other  is  a  circle;  and  they  each  contain  an  acre:  how  much 
further  is  it  around  one  than  the  other? 

Ans.  17.268  yards. 

29.  There  are  two  circles;  the  diameter  of  the  first  is  18 
inches,  and  the  area  of  the  second  is  2-J  times  tliat  of  the 
first:  what  is  the  diameter  of  the  second? 

Ans.  30  inches. 

30.  The  circumference  of  a  circle  is  78.54  inches:  find 
(1)  its  diameter,  and  (2)  its  area. 

Ans.  (1)  25  ins.;  (2)  490.875  sq.  ins. 

31.  A  circle  and  a  square  have  each  a  perimeter  of  120 
feet :  which  contains  the  greater  area,  and  how  much  ? 

Ans.  the  circle,  245.95  sq.  ft. 

32.  What  is  the  width  of  a  ring  between  two  concentric 
circumferences  whose  lengths  are  160  feet  and  80  feet? 

Ans.  12.732  feet. 

33.  Find  the  side  of  a  square  equivalent  to  a  circle  whose 
radius  is  40  feet. 

34.  The  radius  of  a  circle  is  15  feet;  find  the  radius  of  a 
circle  just  three  times  as  large. 


346  PLANE  GEOMETHY. 

35.  The  area  of  a  square  is  225  square  feet:  find  the  area 
of  the  inscribed  circle. 

36.  The  length  of  an  arc  of  180"  is  ttE,  where  E  is  the 
radius  of  the  circle  (436) :  find  the  leugth  of  the  arc  of 
25""  45'  in  the  circle  whose  radius  is  9  inches.     Ans,  4  ins. 

37.  Find  the  number  of  degrees  in  the  arc  whose  length 
is  equal  to  the  radius  of  the  circle.      Ans.  57°  17'  44".8. 

38.  Find  the  number  of  degrees  in  the  arc  whose  length 
is'18  inches,  the  radius  being  5  feet.      Ans.  17°  11'  19". 

39.  Find  the  length  of  the  arc  of  75°  in  the  circle  whose 
radius  is  5  feet.  Ans,  6.545  feet. 

40.  Find  a  side  of  the  circumscribed  equilateral  triangle, 
the  radius  of  the  circle  being  K.  Ans.  2R  V3. 

41.  Find  a  side  of  the  circumscribed  regular  hexagon, 

2T? 

the  radius  of  the  circle  being  R.  Ans. . 

VI 
43.  Find  the  length  of  the  arc  subtended  by  one  side  of 
a  regular  dodecagon  in  a  circle  whose  radius  is  12.5  feet. 

Ans.  6.545  feet. 

43.  Find  the  area  of  a  sector  of  60°  in  the  circle  whose 
radius  is  10  inches.  Ans.  52.3599  sq.  ins. 

44.  The  area  of  a  given  sector  is  equal  to  the  square  con- 
structed on  the  radius:  find  the  number  of  degrees  in  the 
arc  of  the  sector.  Ans.  114°  35'  29".6. 

45.  Find  the  area  of  the  segment  of  60°  in  the  circle 
whose  radius  is  2  feet.  Ans.  0.362344  sq.  ft. 

46.  Find  the  radius  of  the  circle  in  which  the  sector  of 
45°  is  .125  square  inches.  Ans.  .564  inches. 

47.  Two  tangents  make  with  each  other  an  angle  of 
60°:  required  the  lengths  of  the  arcs  into  which  their  points 
of  contact  divide  the  circle,  the  radius  being  21  inches. 

Ans.  44  inches,  88  inches. 

48.  Venice  is  due  south  of  Leipsic  5°  55':  how  many- 
miles  apart  are  they,  the  radius  of  the  earth  being  4000 
miles?  Ans.  413  miles. 

49.  The  three  sides  of  a  triangle  are  9,  10,  17  inches 


BOOK  v.— EXERCISES  IN  MAXIMA  AND  MINIMA.    247 

respectively;  find  (1)  its  area,  and  (2)  the  area  of  the  in- 
scribed circle.  Aiis.  (1)  36  sq.  ins.;  (2)  4;r, 

See  (398),  Ex.  3. 

50.  Mount  Washington  is  visible  from  a  point  at  sea  87 
miles  off:  required  the  height  of  the  mountain. 

Ans.  6270  feet. 

Problems. 

51.  To  circumscribe  a  square  about  a  given  circle. 

52.  To  inscribe  a  regular  octagon  in  a  given  square. 

53.  Given  the  base  of  a  triangle,  the  difference  of  its 
sides,  and  the  radius  of  the  inscribed  circle:  construct  the 
triangle. 

From  the  given  base  AB,  cut  off  AD  =  given  difference;  bisect  DB  in  E,  draw 
EO  ±  to  AB  and  =  the  given  rad.,  etc. 

54.  Describe  a  circle  which  shall  touch  a  given  circle  and 
two  given  straight  lines  which  themselves  touch  the  given 
circle. 

55.  In  a  given  circle  inscribe  a  triangle  whose  angles  are 
as  the  numbers  2,  5,  8. 

See  (453). 

56.  To  inscribe  a  regular  hexagon  in  a  given  equilateral 
triangle. 

57.  To  construct  a  circle  equivalent  to  the  sum  of  two 
given  circles. 

58.  In  a  given  equilateral  triangle,  construct  three  equal 
circles  tangent  to  each  other  and  to  the  sides  of  the  tri- 
angle, and  find  the  radius  of  these  circles  in  terms  of  the 
side  of  the  triangle. 

59.  Construct  an  isosceles  triangle  having  each  angle  at 
the  base  double  the  third  angle. 

60.  Given  the  vertical  angle  of  a  triangle  and  the  radius 
of  the  circumscribed  circle:  find  the  locus  of  the  centre  of 
the  inscribed  circle. 

Describe  a  O  ABC  with  the  given  radius,  draw  AB  cutting  off  segment  ACB 
containing  the  given  Z ;  bisect  arc  AB  opp.  to  C  in  D,  draw  any  chord  DC; 
bisect  ZCAB  by  AO  meeting  CD  in  O,  O  is  the  centre  of  tlie  inscribed  O: 
ZDOA  =  ZOCA  +  zOAC  =  ZBAD  +  ZOAB  =ZDAO;  .-.  DO  =  DA;  ..etc. 

61.  Given  a  vertex  of  a  triangle,  the  circumscribed  cir- 


248  PLANE  GEOMETRY. 

cle  and  the  centre  of  the  inscribed  circle:  construct  the 
triangle. 

Let  A  be  tlie  vertex,  BACD  the  O,  O  the  centre  of  the  inscribed  o  ;  join  AO, 
.produce  it  to  D;  with  centre  D  and  radius  DO  describe  the  o  BOC;  .•.  BD  =  DC 
.-.arc  BD  =  arc  DC;  ZDBO  =  zDOB  =  ZOAB  +  zOBA  =  ZDBC4- ZOBA; 
.-.  zOBC  =  ZOBA,and  ZOAB  =  ZOAC;    .-.etc. 


ExEECisEs  IN  Maxima  akd  Mii^ima. 

62.  Two  sides  of  a  triangle  are  given  in  length:  how 
must  they  be  placed  that  the  area  of  the  triangle  may  be  a 
maximum  ? 

(465). 

63.  Given  the  base  and  vertical  angle  of  a  triangle;  con- 
struct it  so  that  its  area  may  be  a  maximum. 

64.  Of  all  triangles  of  given  base  and  area,  the  isosceles 
is  that  which  has  the  least  perimeter. 

(479). 

65.  Divide  a  given  straight  line  so  that  the  rectangle  con- 
tained by  the  two  segments  may  be  a  maximum. 

66.  A  straight  rod  slips  between  two  straight  rulers  at 
riglit  angles  to  each  other:  in  what  position  is  the  rod  when 
the  triangle  formed  by  the  rulers  and  the  rod  is  a  maxi- 
mum ? 

67.  Show  that  the  greatest  rectangle  which  can  be  in- 
scribed in  a  circle  is  a  square. 

QS.  Of  all  triangles  of  given  vertical  angle  and  altitude, 
the  isosceles  is  that  which  has  the  least  area. 

69.  Of  all  rectangles  of  given  area,  the  square  has  the 
least  perimeter. 

70.  Of  all  polygons  having  the  same  number  of  sides  and 
equal  areas,  the  perimeter  of  an  equilateral  polygon  is  a 
minimum. 

71.  A  and  B  are  two  fixed  points  without  a  circle:  find  a 
point  P  on  the  circumference  such  that  the  sum  of  the 
squares  on  AP,  BP  may  be  a  minimum. 

(333). 


BOOK  V.-EXERC1SE8  IN  MAXIMA  AND  MINIMA.      249 

72.  A  bridge  consists  of  three  arches,  whose  spans  are 
49  ft.,  32  ft.,  and  49  ft.  respectively:  show  that  the  point 
on  either  bank  of  the  river  at  which  the  middle  arch  sub- 
tends the  greatest  angle  is  63  feet  distant  from  the  bridge. 

See  (477). 

73.  If  the  sum  of  the  squares  of  two  lines  be  given,  their 
sum  is  a  maximum  when  the  lines  are  equal. 

74.  Of  all  triangles  having  the  same  base  and  vertical 
angle,  the  isosceles  triangle  has  the  sum  of  the  sides  a  max- 
imum. 

75.  Of  all  triangles  inscribed  in  a  circle,  the  equilateral 
triangle  has  the  maximum  perimeter. 


SOLID  GEOMETRY. 


Book  VI. 
PLANES  AND  SOLID  ANGLES. 

Definitions. 

480.  K  plane  has  been  defined  (11)  as  a  surface  in  which 
the  straight  line  joining  any  two  of  its  points  lies  wholly 
in  the  surface. 

A  plane  is  indefinite  in  extent,  so  that,  however  far  the 
straight  line  is  produced,  all  its  points  lie  in  the  plane. 
But  to  represent  a  plane  in  a  diagram,  it  is  necessary  to 
take  a  limited  portion  of  it,  and  it  is  usually  represented  by 
a  parallelogram  supposed  to  lie  in  the  plane. 

481.  A  plane  is  said  to  be  determined  by  certain  lines 
or  points,  when  it  is  the  only  plane  which  contains  these 
lines  or  points. 

482.  Any  number  of  planes  may  he  passed  through  any 
given  straight  line. 

For,  if  a  plane  is  passed  through 
any  given  straight  line  AB,  the  plane 
may  be  turned  about  AB  as  an  axis, 
and  made  to  occupy  an  infinite  num-      /f| 
ber  of  positions,  each  of  which  will  be       l^ 
a  different  plane  passing  through  AB. 

Hence  a  given  straight  line  does  not  determine  the  posi- 
tion of  a  plane. 

483.  A  plane  is  determined  hy  a  straight  Ime  and  a 
point  without  that  line. 

For,   if    the   plane    containing    the  /      "    .q 

straight  line  AB,  turn  about  this  line  as       / 
an  axis  until  it  contains  the  given  point      I  ^ 
C,  the  plane   is  evidently   determined. 

250 


^ 


1/ 

B 


7 


BOOK  VI.— DEFINITIONS.  251 

If  it  is  then  turned,  in  either  direction,  about  AB,  it  will 
no  longer  contain  the  point  C. 

484.  A  plane  is  determined  hy  three  points  not  in  the 
same  straight  line. 

For,  if  we  join  any  two  of  the  points  by  a  straight  line, 
this  line  and  the  third  point  determine  a  plane.  (483) 

485.  A  plane  is  determined  by  tivo  intersecting  straight 
lines. 

For,  a  plane  passing  through  AB  and 
any  point  0  in  AC,  in  addition  to  the 
point  of  intersection  A,  contains  the 
two  straight  lines  AB,  AC  (480),  and  is 
determined.  (483) 

486.  A  plane  is  determined  by  two  parallel  straight 
lines. 

For,  two  parallel  straight  lines  lie  in 
the  same  plane  (68);  and  since  this 
plane  contains  either  of  these  parallels 
and  any  point  in  the  other,  it  is  deter- 
mined. (483) 

487.  A  straight  line  \^  perpendicular  to  a  plane  when  it  I 
is  perpendicular  to  every  straight  line  which,  it  meets  in ' 
that  plane. 

Conversely,  the  plane  in  this  case  is  said  to  be  perpendic- 
ular to  the  line. 

The  point  in  which  a  line  meets  a  plane  is  called  the  foot 
of  the  line, 

488.  A  straight  line  is  parallel  to  a  jolane  when  it  never 
meets  the  plane,  however  far  both  may  be  produced. 

Conversely,  the  plane  in  this  case  is  said  to  be  parallel  to 
the  line. 

489.  Two  planes  are  parallel  when  they  do  not  meet, 
however  far  they  may  be  produced. 

490.  The  projection  of  a  point  on  a  plane  is  the  foot  of 
the  perpendicular  let  fall  from  the  point  to  the  plane. 


252  SOLID  GEOMETRY. 

491.  The  projection  of  a  line  on  a  plane  is  the  locus  of 
the  projections  of  all  the  points  of  this  line. 

492.  The  a7igle  which  a  straight  line  makes  with  a  plane 
is  defined  to  be  the  acute  angle  between  the  straight  line 
and  its  projection  on  the  plane,  and  is  called  the  inclina- 
tion of  the  line  to  the  plane. 

493.  By  the  dista^ice  of  a  point  from  a  plane  is  meant 
the  shortest  distance  from  the  poitit  to  the  plane. 

494.  The  line  that  determines  the  projection  of  a  point 
on  a  plane  is  called  tha  2)roJectin(/  line  of  that  point.  The 
jilane  including  all  the  projecting  lines  of  a  straight  line  is 
called  the  projecting  plane  of  the  line. 


Lines  and  Planes. 

Proposition  1 .    Tiieorem. 

495.  If  two  planes  cut  each  other,  their  common  inter- 
section is  a  straight  line. 

Hyp,    Let  AB,  CD   be    two 
planes  which  cut  each  other. 

To  prove  their  common  inter- 
section is  a  straight  line. 

Proof.  Let  E  and   H  be  any 
two  pts.  common  to  both  planes. 

Join  E  and  H  by  the  st.  line 
EH. 

Because  E  and  H  are  in  both 
planes,  the  st.  line  EH  lies  in  both  planes.  (480) 

Because  a  st.  line  and  a  pt.  out  of  it  cannot  lie  in  two 
planes,  {'^^^) 

.  • .  EH  contains  all  the  pts.  common  to  both  planes. 
.  • .  EH  is  the  common  intersection  of  the  two  planes. 

Q.E.D. 


I 


BOOK   VL-LINE8  AND  PLANES.  253 


Proposition  2.    Tiieorem. 

496.  If  oUique  lines  are  draiofi  from  a  i)oint  to  a  plane: 
■  (1)  Two  oUiqiie  lines  meeting  the  plane  at  equal  distances 
from  the  foot  of  the  perpendicular  are  equah 

(2)  Of  two  oblique  lines  meeting  the  plane  at  unequal  dis- 
tances from  the  foot  of  the  perpendicular,  the  7nore  remote 
is  the  longer. 

Hyp.  Let  OP  be  _L  to  the  plane 
MN, 

and  PA  =  PB, 

and  PC  >  PA. 

To  prove  (1)  OA  =  OB, 

and  (2)  00  >  OA. 

(1)  Proof     Since  OP   is  com- 
mon, 

and  PA  =  PB,  (Hyp.) 

.-.  rt.  A  OPA  =  rt.  A  OPB.  (104) 

.•.OA=rOB. 

^  (2)  Proof    On  PC  take  PB  =  PA,  and  join  OB. 

*  Then,  OC  >  OB.  (63) 

But  OB  =  OA.  (Just  proved) 

.-.  0C>  OA.  •    Q.E.D 

497.  Cor.  1.  The  perpendicular  is  the  shortest  distance 
from  a  p)oint  to  a  plane;  therefore,  hy  the  distance  of  a 
point  from  a  plane  is  7neafit  the  peipendicular  distance 
from  the  point  to  the  plane.  (493) 

498.  Cor.  2.  Equal  oblique  lines  from  a  point  to  a  plane 
meet  the  plane  at  equal  distances  from  the  foot  of  the  per- 
pendicular ; .  and  of  two  unequal  oblique  lines,  the  greater 
meets  the  plane  at  the  greater  distance  from  the  foot  of  the 
perpendicular. 

499.  Cor.  3.  The  locus  of  the  point  in  a  2^la7ie  at  a  given 
distance  from  a  fixed  2Joint  loithout  the  plane,  is  a  circle 
whose  centre  is  the  foot  of  the  perpendicular. 


254 


SOLID   GEOMETRY. 


o 

"rv 

\^  ^N 

\\\ 

M, 

/ 

\\^^l 

/ 

/     P 

5«rn:^rr_47''' /'       / 

/       'X'V^''       / 

/              /A/  /'         / 

/           1     /  /  /           / 

i  ///         ^ 

(//'' 

K 

0 

Nf 


Proposition  3.    TFieorem. 

^  500.  7/*^  straight  line  is  jJ^^pendicula?'  to  each  of  tioo 
straight  lines  at  their  point  of  intersection,  it  is  perpenr 
dicidar  to  the  plane  of  those  lines. 

Hyp.  Let  OP  be  J.  to  PA,  PB 
at  the  pt.  P. 

To  prove  OP  is  _L  to  the  plane 
MN  of  these  lines. 

Proof.  Join  AB,  and  through  P 
draw  in  MN  any  other  st.  line  PC 
cutting  AB  in  C. 

Produce  OP  to  0'  making  PO'  =: 
PO,  and  join  0,  0'  to  each  of  the 
pts.  A,  B,  C. 

Since  PA,  PB  are  I.  to  00' at  its  mid.  pt.,  (Hyp.)  (Cons.) 

.• .  OA  =  O'A,  and  OB  =  O'B.  (66) 

.• .  A  OAB  =  A  O'AB,  and  .• .  z  OAC  =  Z O'AC.      (108) 

Also,  A  OAC  =  A  O'AC, 

having  two  sides  and  ilie  included  Z  equal  each  to  each  (104).        , 
.  • .  OC  =  O'C,  and  .  • .  PC  is  ±  to  00'  at  its  mid.  pt.  P.  (67) 

.* .  OP  is  JL  to  any  st.  line  in  MN  passing  through  its  foot  P. 
.  • .  OP  is  J.  to  the  plane  MN.  (487) 

Q.E.D. 

]  501.  Cor.  \.  At  a  given  point  in  a  plane,  only  one 
perpendicular  to  the  plane  can  he  erected. 

For,  if  there  coukl  be  two  J_s  at  the  same  pt.  P,  pass  a 
plane  through  them  whoso  intersection  with  the  plane  MN 
is  AP;  then  these  two  J_s  would  be  both  J_  to  the  line  AP 
at  the  same  pt.  P,  which  is  impossible.  (51) 

502.  Cor.  2.  From  a  given  point  without  a  plane  only 
one  perpendicular  can  he  draica  to  the  plane. 

For,  if  OP,  OA  be  two  sue]]  _Ls,  the  A  OPA  contains  two 
rt.  Z  s,  which  is  impossible.  (1^1) 


y 


BOOK  VI. -PERPENDICULARS  TO  LINES.        255 


Proposition  4.    Theorem. 


503.  Cot^veri>6ly,  all  the  i^erpendicidars  to  a  straight 
line  at  the  same  point  lie  in  a  plane  perpendicular  to  the 
line. 

Hyp,  Let  PA,  PB  be  two  st.  lines 
J_  to  OP  at  P,  and  PC  any  other 
line  _L  to  OP  at  P. 

To  prove  PC  is  in  the  same  plane 
with  PA,  PB. 

Proof.  Let  the  plane  passing 
through  PO  and  PC  cut  the  plane 
APB  in  the  line  PC. 

Then  OP  is  _L  to  PC.  (487) 

But  in  the  plane  OPC  only  one  _L  can  be  drawn  to  OP 
atP,  (51) 

.-.PC  and  PC  coincide,  and  PC  lies  in  the  plane  APB. 

Q.E.D. 

Note.— Hence  a  plane  is  determined  by  one  point  and  the  normal  to  the  plane 
at  that  point. 

504.  Cor.  1.  At  a  given  point  in  a  straight  line  one 
plane,  and  only  one,  can  he  drawn  j)erpendicular  to  the 
li7ie. 

505.  Cor.  2.  If  a  right  angle  he  turtied  roimd  one  of 
its  arms  as  an  axis,  the  other  arm  ivill  generate  a  plane. 

506.  Cor.  3.  Tlirough  a  given  point  luithont  a  straight 
line  one  plane,  and  only  one,  can  he  drawn  perpe7idicular 
to  the  line. 

For,  in  the  plane  of  OP  and  the  pt.  C,  the  J^  CP  can  be 
drawn  to  OP;  then  the  piano  genei-ated  by  turning  PC 
round  OP  will  be  J_  to  OP;  and  it  is  clear  that  there  is 
only  one  such  J_  plane. 


256 


SOLID  GEOMETRY. 


Proposition  5.    Theorem. 

507.  If  from  the  foot  of  a  perpendicular  to  a  plane  a 
straight  line  is  drawn  at  right  angles  to  anfline  in  the 
plane,  and  its  intersection  with  that  line  is  joined  to  any 
point  of  the  perpendicular,  this  last  line  will  be  per2)en- 
diciilar  to  the  line  in  the  plane. 

Hyp.  Let  OP  be  a  _L  to  the  plane 
MN,  PA  a  J_  from  P  to  any  line  BC 
in  MN,  and  OA  a  line  joining  A  with 
any  pt.  0  in  OP. 

To  prom         that  OA  is  ±  to  BC. 

Proof.  Take  AB  =  AC,  and  join 
PB,  PC,  OB,  OC. 

Since  PA  is  _L  to  BC  at  its  mid.  pt., 


and 


PB  =  PC, 
OB  =  OC. 


(496) 


Then  since  0  and  A  are  each  equally  distant  from  B 
and  C, 

.-.  OA  is_LtoBC.  (67) 

Q.E.D. 

508.  Cor.  1.   Tlie  line  BC  is perpendictdar  to  the  plane 

of  the  triangle  OP  A . 

For,  it  is  i.  to  the  st.  lines  AP,  AO  at  pt.  A.  (500) 

Cor.  2.   The  line  PA  measures  the  shortest  distance 

hetiveen  OP  and  BC. 


EXERCISES. 

1.  Find  the  locus  of  points  equally  distant  from  two 
given  points. 

2.  Given  a  straight  line  and  any  two  points:  find  a  point 
in  the  straight  line  equally  distant  from  the  two  points. 


BOOK  VL—PERPENDICULAR  AND  PARALLEL  LINES.  257 


Proposition  6.    Theorem. 

n/     509.  Tivo  straight  lines  perpendicular  to  the  same  plane 
are  parallel. 

Hyp.  Let  AB;  CD,  be  _L  to  the 
plane  MN  at  the  pts.  B,  D.  j^ 

To  prove  AB  ||  to  CD. 

Proof.  Join   DB,  DA,  and  ^w 
DE  J_  to  BD  in  t(ie  plane  MN. 

Then  CD  is  JL  to  ED, 

and  BD  is  J.  to  ED, 

and  AD  is  _L  to  ED. 

.  •.  CD,  AD,  BD,  are  all  in  the  same  plane, 

all  the  Is  to  a  si.  line  at  the  samepL  lie  in  the  same  plane  (503). 

.  • .  AB  and  CD  lie  in  the  same  plane. 


they  are  both  JL  to  BD. 
.-.  AB  is  II  to  CD. 


and  thev  are  both  J    to  BD.  (487) 

(70) 

Q.E.D. 

510.  Cor.  1.  If  one  of  two  parallel  lines  is  perpen- 
dicular to  a  plane,  the  other  is  also  perpendicular  to  that 
plane.  A       c 

For,  if  AB  is  jj  to  CD,  and  J_  to  the  j^ 
plane  MN,  then  a  J.  to  MN  at  D  will  be  jj         / 

to  AB   (509),  and  will  coincide  with  CD  / 

(501).     .  • .  CD  is  _L  to  MN.  ^N 

511.  Cor.  2.    Two  straight  lines  that  are  parallel  te  a 
third  straight  line  are  fjarallel  to  each  APE 
other. 

For,  each  of  the  lines  AB,  CD,  is  J_ 
lo  a  plane  MN  that  is  _L  to  EF  (510);         y 
.  • .  AB  and  CD  are  I| .  (509)      N^ 


7 


U/ 


M 


258 


SOLID  OEOMETRT. 


Parallel  Planes. 
Proposition  7.    Theorem. 

512.  If  two  straiglit  lines  a7'e  parallel,  each  of  them  is 
parallel  to  every  plane  passing  through  the  other  and  not 
containing  loth  lines. 

Hyp,  Let  AB,  CD  be  two  ||  st.  lines, 
and  MN  any  plane  passing  through 
CD. 

To  prove        AB  ||  to  MN". 

Proof  The  ||  s  AB,  CD  lie  wholly 
in  the  plane  ABCD. 

.  • .  if  AB  meets  the  plane  MN,  it  mnst  meet  it  in  some 
pt.  of  the  intersection  CD,  of  the  two  planes. 

But  AB  is  II  to  CD,  and  so  cannot  meet  it.  (HyP-) 

.  • .  AB  cannot  meet  the  plane  MN. 

.-.  AB  is  II  to  MN.  Q.E.D. 

613.  Cor.  1.  A  line  parallel  to  the  intersection  of  tivo 
planes  is  parallel  to  each  of  those  planes . 

514.  Cor.  2.  Through  any  given  straight  line,  a  plane 
can  be  passed  j^arallel  to  any  other  given  straight  line. 

For,  through  any  pt.  C  of  CE  draw 
CD  II  toAB;  then  the  plane  of  DCE 
is  II  to  AB.  (512) 

515.  Cor.  3.  Tlirough  a  given  point 
a  plane  can  he  passed  parallel  to  any 
two  given  straight  lines  in  space. 

For,  draw  through  the  given  pt.  0, 
in  the  plane  of  the  given  line  AB  and 
O,  the  line  A'B'  i|  to  AB,  and  in  the 
plane  of  the  given  line  CD  and  0,  the 
line  CD'  ||  to  CD;  then  the  plane  of 
the  lines  A'B',  CD'  is  ||  to  each  of 
the  lines  AB  and  CD, 


BOOK   VL— PARALLEL  PLANES. 


259 


Proposition  8.    Theorem. 

516.  Two  planes  perpendicular  to  the  same  straight  line 
are  pitrallel. 


^c 


Hyp.  Let  the  planes  MN  and  PQ  be 
AB. 


0  prove 


MN  II  to  PQ. 


Proof.  If  the  planes  are  not  jj  they 
will  meet  if  sufficiently  produced.  (489) 

There  will  then  be,  through  a  pt.  of 
their  intersection,  two  planes  J_  to  the 
same  st.  line  AB. 


/J 

p 

N 

/  = 

../ 

But  this  is  impossible. 

.-.MN  is  II  to  PQ. 


(506) 

Q.E.D. 


517.  Cor.  1.  If  a  straight  line  is  parallel  to  a  plane, 
the  intersection  of  the  plane  witJi  any  plane  passed  through 
the  line  is  parallel  to  the  line. 

Let  the  student  show  this. 

518.  Cor.  2.  If  a  straight  line  and  a  plane  are  parallel, 
a  parallel  to  the  line  drawn  through  any  point  of  the  plane 
lies  in  the  plane,  (517) 

Hem.  a  polygon  in  space  may  be  formed  by  joining  end 

to  end  any  number  of  straight  lines,  as  defined  in  (137). 

,But  in  Plane  Geometry  the  lines  are  all  confined  to  one 

plane,  while  there  is  no  such  restriction  upon  polygons  in 

space. 


260 


SOLID  GEOMETRY, 


Proposition  9.    Tiieorem. 

519.  The  intersections  of  two  parallel  planes  hy 
plane  are  parallel  lines. 

Hyp.  Let  the  li  planes  MN,  PQ  be 
cut  by  the  plane  AD  into  the  lines  AB 
and  CD. 

To  prove  AB  ||  to  CD. 

Proof,  The  lines  AB,  CD  are  in  the 
same  plane  AD. 

Because  AB  lies  in  the  plane  MN, 
and  CD  in  the  I|  plane  PQ,  and  because 
these  planes  cannot  meet, 

.• .  AB  and  CD  cannot  meet. 


a  third 


AB  is  II  to  CD. 


Q.E.D. 


520.  Cor.  Parallel  straight  lines  included  between  pai*- 
allel  planes  are  equal. 

For,  the  plane  of  the  ||  lines  AC,  BD  intersects  the  I| 
planes  MN,  PQ  in  the  ||  lines  AB,  CD. 

.• .  ABCD  is  a  O,  and  .• .  AC  =  BD.  (129) 


EXERCISES. 

1.  Show  that  all  the  propositions  in  Plane  Geometry 
which  relate  to  triangles  are  true  of  triangles  in  space,  how- 
ever situated. 

See  (485).      ^c^u     , 

2.  Show  that  those  propositions  are  not  true  of  polygons 
of  more  than  three  sides  situated  in  any  way  in  space. 


D 


BOOK  VI.— PARALLEL  PLANES.  261 

Proposition  lO.    Theorem. 

521.  ^  straight  line  perpendicular  to  one  of  two  parallel 
planes  is  j)erpendicular  to  the  other. 

HifiJ.  Let  MN  and  PQ  be  ||  planes,      "^y 7 

and  "let  AB  be  _L  to  PQ.  /       A. C  / 

To  prove  AB  ±  to  MK  I- '- 

Proof.   Through  A  draw  any  st.  line        p 
AC  in  the  plane  MN",  and  pass  a  plane       / 

through  AB,  AC,  intersecting  PQ  in     ^ 

BD. 

Then  AC  is  ||  to  BD.  (519) 

But  AB  is  J_  to  BD.  (487) 

.-.  AB  is  Jl  to  AC.  (71) 

And  since  AC  is  any  line  drawn  through  A  in  the  plane 
MN, 

/.  AB  is  i.  to  the  plane  MK  (487) 

Q.E.D. 

522.  CoK.  1.  Through  a  given  point  one  plane  can  he 
passed  parallel  to  a  given  2^lane,  and  only  one. 

For,  if  AB  is  J_  to  PQ,  a  plane  passing  through  the  pt. 
A,  J.  to  AB,  is  II  to  PQ.  (51G) 

Also,  since  every  plane  ||  to  PQ  is  _L  to  AB  (521),  and 
since  only  one  plane  can  be  passed  through  the  pt.  A  J.  to  AB 
(504),  therefore  only  one  plane  can  be  passed  through  a 
given  pt.  parallel  to  a  given  plane. 

523.  CoK.  2.  Two  parallel  j^lctnes  are  everywhere  equally 
distant. 

For,  all  st.  lines  J_  to  the  plane  PQ  are  also  J_  to  the  |i 

plane  MN  (521) ;  and  being  J_  to  the  same  plane,  they  are 

II  (509)  ;  and  being  included  between  ||  planes,   they  are 

equal   (520).     Hence   the  planes   are   everywhere  equally 

distant.  *  (497) 

524.  Cor.  3.  If  tioo  intersecting  straight  lines  are  each 
parallel  to  a  given  plane,  the  plane  of  these  lines  is  parallel 
to  the  given  plane. 

See  (518),  (516). 


262 


SOLID  OEOMETRY. 


Ur 


Proposition  1  1 .    Theorem. 

525.  If  two  angles  not  in  the  same  j^lane  have  their  sides 
respectively  parallel  and  lying  in  the  same  direction,  they 
are  equal  and  their  planes  are  parallel. 

Hyp,  Let  Zs  A  and  A'  lie  in  the 
planes  MN,  PQ  respectively,  and  let 
AB  be  II  to  A'B'  and  AC  be  ||  to  A'C. 

(1)  To  prove         ZA=  /A'. 

Proof,  Take  AB  =  A'B',  and  AC  = 
A'C,  and  join  AA',  BB',  CC,  BC, 
B'C.  ; 

Then,  since        AB  is  =  and  ||  to  A'B', 

.-.  AA'  is  =  and  ||  to  BB'. 

In  like  manner 

A  A'  is  =  and  ||  to  CC. 

Because  BB'  and  CC  are  each  =  and  ||  to  AA', 

.-.  BB'  is  =  and  ||  to  CC.  (511) 

.-.  BC  is  =  and  ||  to  B'C.  (133) 

.-.  A  ABC  =  aA'B'C, 
having  tlie  three  sides  equal  each  to  each  (108). 

.-.  ZA=ZA'. 

(2)  Tojyrove  MN  |I  PQ. 

Proof,    Since  the  lines  AB,  AC  are  each  ||  to  the  plane 

PQ,      '  (512) 

.-.  their  plane  MN  is  ||  to  PQ.  '  (524) 

Q.E.D. 
EXERCISE. 

Find  the  locus  of  points  equally  distant  from  three  given 
points. 


BOOK  VL— PARALLEL  PLANES. 


263 


Proposition  12.    Theorem. 

526.  If  two  straight   lines   are   cut  by  three  parallel 
planes,  the  corresponding  segmeyits  are  proportional. 

Hyp.  Let  AB,  CD  be  cut  by  the  ||     U^ _^ 

planes  MN,  PQ,  RS,  in  the  pts.  A,  E,  B,     /Kr'-^f       / 
and  C,  F,  D.  ^" 


To  prove 


EB 


CP 
FD- 


isn 


M 


Proof.  Join  AD  cutting    the   plane    / 
PQ  in  H.  ^ 

Join  EH  and  FH. 

Then  because  the  ||  planes  PQ,  RS  are  cut  by  the  plane 
ABD,  in  the  lines  EH,  BD, 


.-.  EH  is  II  to  BD. 
.-.  AE:EB  =  AH:HD. 
In  like  manner        HF  is  ||  to  AC 

.-.  AH:HD  =  CF:FD. 
.-.  AE:EB  =  CF:FD. 


(519) 
(298) 
(519) 

Q.E.D. 


EXERCISES. 

1.  To  draw  a  perpendicular  to  a  given  plane  from  a 
given  point  without  it. 

2.  To  erect  a  perpendicular  to  a  given  plane  from  a  giv- 
en point  in  the  plane. 

3.  Prove  that  through  a  given  line  of  a  given  plane, 
only  one  plane  perpendicular  to  the  given  plane  can  be 
passed. 


264  SOLID  GEOMETRY. 


DiEDRAL  Angles. 

DEFINITIONS. 

527.  When  two  planes  intersect  they  are  said  to  form 
with  each  otlier  a  diedral  angle. 

The  two  planes  are  called  the  faces,  and  their  line  of  in- 
tersection, the  edf/ey  of  the  diedral  angle. 

Thus,  the  two  planes  AC,  AE  are  the 
faces,  and  the  intersection  AB  is  the  edge, 
of  the  diedral  angle  formed  by  these 
planes. 

A  diedral  angle  is  read  by  the  two  let- 
ters on  the  edge  and  one  in  each  face,  the 
two   on  the  edge  being  read  between  the  other  two;  or, 
simply  by  the  two  letters  on  the  edge. 

Thus,  the  angle  in  the  figure  is  read  either  DABF  or 
AB. 

528.  If  a  point  is  taken  in  the  edge  of  a  diedral  angle, 
and  two  straight  lines  are  drawn  through"  this  point,  one  in 
each  face,  and  each  perpendicular  to  the  edge,  the  angle 
between  these  lines  is  called  the  plaiie  angle  of  the  diedral 
angle.  This  plane  angle  is  the  same  at  whatever  point  of 
the  edge  it  is  constructed.  '' 

Thus,  if  at  any  point  G  we  draw  GH  and  GK  in  the  two 
faces  AC  and  AE  respectively,  and  both  perpendicular  to 
AB,  the  angle  HGK  is  equal  to  the  angle  DAF,  since  the 
sides  of  these  angles  are  respectively  parallel.  (525) 

The  plane  of  the  plane  angle  HGK  is  perpendicular  to 
the  edge  AB  (500);  and  conversely,  a  plane  perpendicular 
to  the  edge  of  a  diedral  angle  at  any  point  cuts  the  faces  in 
lines  perpendicular  to  the  edge.  (487) 


BOOK  VI.—DIEDRAL  ANGLES. 


265 


529.  A  diedral  angle  may  be  conceived  to  be  generated 
by  revolving  a  plane  about  a  line  of  the  plane. 

Thus,  suppose  a  plane,  at  first  in  coincidence 
with  a  fixed  plane  AC,  to  turn  about  the  edge 
AB  as  an  axis  until  it  comes  into  the  position 
AE;  then  the  magnitude  of  the  diedral  angle 
DABF  varies  continuously  with  the  amount  of 
turning  of  tliis  plane  about  AB.  Tlie  straight 
line  DA,  perpendicular  to  AB,  generates  the  plane  angle 
DAF. 

530.  Two  diedral  angles  are  equal  when  one  of  them 
can  be  applied  to  the  other  so  that  the  edges  coincide,  and 
the  two  plane  faces  of  the  one  coincide  respectively  with  the 
two  plane  faces  of  the  other. 

The  magnitude  of  a  diedral  angle  depends  only  upon 

the  relative  position  of  its  faces,  and  is  independent  of  their 

extent. 

A__ ^c 


531.  When  two  diedral  angles  have  a  com- 
mon edge,  and  the  intermediate  face  common 
to  both,  they  are  said  to  be  adjacent. 

Thus,  the  angles  CABD,  DABE  are  adja- 
cent angles. 

Two  diedral  angles  CABD,  DABE,  are  added  together 
by  placing  them  adjacent  to  each  other,  giving  as  the  sum 
the  diedral  angle  CABE. 

532.  When  the  adjacent  diedral  angles  which  a  plane 
forms  with  another  plane  on  opposite  sides  are  equal,  each 
of  these  angles  is  called  a  right  diedral  angle  ;  and  the  first 
plane  is  said  to  be  jjerpendicular  to  the  other. 

Thus,  if  the  adjacent  diedral  angles 
ABCM,  ABClSr  are  equal,  each  of  these  is 
a  right  diedral  Jicgle,  and  the  planes  AC 
and  M"N  are  perpendicular  to  each  other. 

Through  a  given  line  in  a  plane  only 
one  plane  can  be  passed  perpendicular  to 
the  given  plane. 


E/1 

A 


M 


7" 


266 


SOLID  GEOMETMT, 


Proposition  13.    Theorem. 

533.   Two  diedral  a7igles  are  equal  if  their  plane  angles 
are  equal. 


Hyp.  Let  the  plane  /  s  CAD,    A 
C'A'D'   of   the   diedral  /s  AB, 
A'B'  be  equal. 

To  prove 

diedral  Z  AB  =  diedral  Z  A'B'.   B 


C.A" 


B' 


C 


Proof,  Apply  the  diedral  Z  AB  to  the  diedral  ZA'B',  so 
that  the  plane  ZCAD  coincides  with  the  equal  plane 
Z  C'A'D'. 

Because  the  pt.  A  coincides  with  the  pt.  A',  and  the 
plane  CAD  with  the  plane  C'A'D',  (485) 

.*.  AB,  the  _L  to  the  plane  CAD,  coincides  with  A'B', 
the  J.  to  the  plane  C'A'D'.  (501) 

Because  AB  coincides  with  A'B',  and  AC  with  A'C, 

.'.  the  plane  BC  coincides  with  the  plane  B'C.       (485) 

Similarly,  the  plane  BD  coincides  with  the  plane  B'D'. 

.-.  diedral  Z  AB  =  diedral  Z  A'B'.  (530) 

Q.E.D. 

534.  ScH.  A  diedral  angle  is  called  acute,  right,  or  ob- 
tuse according  as  the  corresponding  plane  angle  is  acute, 
right,  or  obtuse. 

EXERCISE. 

Prove  that  through  a  line  parallel  to  a  given  plane;  only 
one  plane  perpendicular  to  the  given  plane  can  be  passed. 


BOOK  VL—DIEDRAL  ANGLES. 


267 


Proposition  14.    Theorem. 

535.  The  ratio  of  any  two  diedral  aiigles  is  equal  to  the 
ratio  of  their  i^lane  angles. 


Htjp.  Let  CABD,  C'A'B'D'  be  two  diedral  Zs,  and  let 

CAD,  C'A'D'  be  their  plane  Z  s. 

AB       ZCAD 
Toprove  _^,^__,__. 

Proof.  Take  any  common  measure  of  the  Zs  CAD, 
C'A'D',  and  suppose  it  to  be  contained  three  times  in  Z  CAD 
and  4  times  in  Z  C'A'D'. 

Then  Z  CAD  :  Z  C'A'D'  =  3:4. 

Pass  planes  through  the  edges  AB,  A'B'  and  the  several 
lines  of  division  of  the  plane  Z  s  CAD,  C'A'D'  made  by 
the  common  measure. 

The^e  planes  divide  CABD  into  3,  and  C'A'B'D'  into  4 
equal  parts.  (533) 

.-.  CABD  :  C'A'B'D' =  3  :  4. 
.-.  CABD  :  C'A'B'D'  =  zCAD  :Z C'A'D'. 

This  proof  may  be  extended  to  the  case  where  the  plane 
angles  are  incommensurable,  by  the  method  employed  in 
(234),  (298),  and  (356).  q.e.d. 

536.  ScH.  Since  the  diedral  angle  and  its  plane  angle 
increase  and  decrease  in  the  same  ratio,  the  plane  angle  is 
taken  as  the  measure  of  the  diedral  angle. 

See  (236). 


268 


SOLID  GEOMETRY. 


Proposition  1  5.    Tl-ieorem. 

537.  If  iioo  planes  are  iderpendicular  to  each  other,  a 
straight  line  drawn  in  one  of  them  2)erpendicular  to  their 
intersection  is  perpendicular  to  the  other. 

Hyp.  Let  the  planes  PQ  and  MN 
be  _L  to  each  other,  and  let  AB  be 
drawn  in  PQ  J_  to  their  intersection 
QC. 

To  prove     AB  _L  to  MN. 

Proof  In  the  plane  MN  draw  BD 
X  to  QC  at  B. 

Because  BA  and  BD  are  each  _L  to 
QC,  (Hyp.)  and  (Cons.) 

.  • .  Z  ABD  is  the  plane  Z  of  the  rt.  diedral  Z  PCQN.  (528) 

Because  PQ  is  _L  to  MN,  (Hyp.) 

.-.ZABDisart.  Z.  (53G) 

.  • .  AB  is  _L  to  QC  and  BD  at  their  pt.  of  intersection. 


.  AB  is  _L  to  the  plane  MN. 


(500) 

Q.E.D. 


538.  Cor.  1.  If  tioo  ^jlanes  are  perpendicular  to  each 
other,  a  straight  line  through  any  j^oint  of  their  intersec- 
tion perpendicular  to  one  of  the  planes  %uill  lie  in  the  other. 

See  (501). 

539.  Cor.  2.  If  tivo  planes  are  perj)e7idicular  to  each 
other,  a  straight  line  from  any  poi7it  of  one  plane  peipefi- 
dicular  to  the  other  will  lie  in  the  first  plane. 

See  (502). 


BOOK    Vl.—DIEDHAL  ANGLES. 


269 


Proposition  1 6.    Theorem. 

540.  If  a  straight  line  is  perpendicular  to  a  plane,  every 
plane  passed  through  the  line  is  perpendicular  to  that 
plane. 

Hyp.  Let  AB  be  _L  to  the  plane 
MN,  and  PQ  any  plane  passed 
through  AB,  intersecting  MN  in  QO. 

To  prove  plane  PQ  J_to  plane  MN. 

Proof.  In  the  plane  MN  draw  BD 
i.  to  QC  at  B. 

Because  AB  is  _L  to  MN,     (Hyp.) 


Q 'n 


.• .  AB  is  J_  to  QC  and  BD.  (487) 

.  • .    Z  ABD  is  the  plane  Z  of  the  diedral  Z  PCQN.     (528) 
But  Z  ABD  is  a  rt.  Z .  (Proved  above) 

.-.  PQis_LtoMN.  Q.E.D. 

541.  Cor.  A  plane  perpendicular  to  the  edge  of  a  diedral 
angle  is  perpendicular  to  its  faces. 

542.  ScH.  When  three  straight  lines,  AB,  CB,  DB,  are 
perpendicular  to  one  another  at  the  same  point,  each  line 
is  perpendicular  to  the  plane  of  the  other  two,  and  the 
three  planes  are  perpendicular  to  one  another. 


EXERCISE. 


If  two  planes  which  intersect  contain  two  lines  parallel 
to  each  other,  the  intersection  of  the  planes  will  be  parallel 
to  the  lines. 


270  SOLID  GEOMETRY. 


Proposition  1  7.    Theorem. 

543.  TJirough  a  given  straight  line  oblique  to  a  plane,  a 
plane  can  he  passed  perpendicular  to  the  given  plane,  and 
hut  one. 

Hyp.  Let  AB  be   the  given  st.  line 
oblique  to  the  plane  MIS'. 

To  prove  that  one  plane  can  be  passed        /"[  j     7 

through  AB  _L  to  MN,  and  but  one.  /  ^^  'D  / 

Proof.  Draw  AD  J_  to  MN  from  any     /  / 

pt.  AofAB.  ^ 

Through  AB  and  AD  pass  a  plane  AC, 

Because  the  plane  AC  passes  through  the  _L  AD^  (Cons.) 

.'.  the  plane  AC  is  _L  to  the  plane  MN.  (540) 

Also,  because  any  plane  passed  through  AB  _L  to  MN" 
must  contain  the  _]_  AD,  (539) 

.  • .  the  plane  AC  is  the  only  plane  JL  to  MN  that  can  be 
passed  through  AB.  (485) 

Q.E.D. 

544.  CoR.  1.  Tlie  projection  of  a  straight  line  on  a  plane 
is  a  straight  line. 

For,  the  J_s  from  all  pts.  of  AB  to  [MN  lie  in  the  plane 
AC  J_  to  MN  (539),  and  therefore  these  J_s  all  meet  MN 
in  the  intersection  CD  of  the  two  planes,  which  is  a  straight 
line.  (495) 

545.  Cor.  2.  If  a  lifie  intersects  a  p)ln'ne,  its  p)rojection 
passes  through  the  point  of  intersection. 


BOOK    Vl.'-rERPENDICULAR  PLANES, 


271 


Proposition  18.    Theorem. 

546.  If  Uvo  intersecting  planes  are  each  perperidicular 
to  a  third  plane,  their  intersection  is  perpendicular  to  the 
third  plane. 

Hyp.  Let  the  planes  PQ,  RS,  in- 
tersecting in  AB,  beJ_to  the  third 
plane  MN. 

To  prove       AB  _L  to  MN. 

Proof,  At  the  pt.  B  erect  a_Lto 
the  plane  MN. 

This  _L  lies  in  each  of  the  planes 
PQ  and  ES. 


Because  this  _L  lies  in  each  plane  PQ  and  RS,  it  is  their 
line  of  intersection  AB. 


.  AB  is  JL  to  the  plane  MN. 


Q.E.D. 


547.  Cor.  1.  A  plane  perpendicular  to  each  of  two  in- 
tersecting planes  is  perpendicular  to  their  intersection, 

548.  Cor.  2.  If  the  planes  PQ  and  RS  include  a  right 
diedral  angle,  the  three  platies  PQ,  ES,  MN,  are  perjjen- 
dicular  to  one  another ;  the  intersection  of  any  two  of 
tliese  planes  is  perpendicular  to  the  third  plane ;  and  the 
three  intersections  are  perpendicular  to  one  another. 

Compare  (54S). 


EXERCISE. 


Show  that  three  planes  in  general  intersect  in  one  point: 
what  are  the  exceptions  ? 


272  SOLID   GEOMETliY. 


Proposition  19.    Theorem. 

549.  Every  i^oint  in  the  2:)lane  bisecting  a  diedral  angle 
is  equally  distant  from  the  faces  of  the  angle, 

Hyi).  Let  AM  be  the  plane 
which  bisects  the  diedral 
Z  CABD,  and  let  PE,  PH  be_Ls 
from  any  pt.  P  in  the  plane  AM 
to  the  planes  CB  and  DB. 

To  prove      PE  =  PH. 

Proof  Let  0  be  the  pt.  where 
the  plane  EPH  cuts  the  line  AB. 

Join  EO,  OP,  OH. 

Because   PE   is_Lto   CB  and 

PH  is  _L  to  DB,  (Cons.) 

.  • .  the  plane  EPH  is  _L  to  each  of  the  planes  CB  and 
DB.  (540) 

.  • .  the  plane  EPH  is  _L  to  their  intersection  AB.    (547) 

.  • .  the  Z  s  POE,  POH  are  the  plane  Z  s  of  the  diedral 
Z  s  CABM  and  DABM.  (528) 

Because  the  diedral  Z  s  CABM,  DABM  are  equal,  (Hyp.) 

.-.  ZP0E=  ZPOH.  (536) 

.-.  rt.  aPOE  =  rt.  A  POH, 

hamng  the  hypotenuse  and  an  acute  Z  =  each  to  ea^h  (106). 

.•.PE  =  PH.  Q.E.D. 

EXERCISES. 

1.  If  three  planes  have  a  common  line  of  intersection, 
the  normals  drawn  to  these  planes  from  any  point  of  that 
line  are  in  one  plane. 

2.  If  from  any  point  perpendiculars  be  drawn  to  the 
faces  of  a  diedral  angle,  the  angle  between  these  perpen- 
diculars is  the  supplement  of  the  diedral  angle  between  the 
planes  in  which  the  point  is  situated. 


BOOK   VI.—LMAST  ANGLE   WITH  A  PLANE.      2T6 


Proposition  20.    Theorem.* 

550.  TliG  acute  angle  between  a  straight  line  and  its 
projection  on  a  plane  is  the  least  angle  which  the  line  makes 
luith  any  line  of  the  plane. 

Hyp.  Let  BO  be  the  projection  of  ^/<l 

the  St.  line  AB  on  the  plane  MM,  and       M '^         l\ 

let  BD  be  any  other  st.  line  in  the       /b<™ i-'ol 

plane,  drawn   through  B,  the  pt.   in      /           ^^^^-'n    / 
which  AB  meets  the  plane.  ^ -^ 

To  jjrove  I  ABC  <  Z  ABD. 

Proof.     Take  BD  =  BO,  and  join  AD. 

Because  in  the  A  s  ABO,  ABD, 

AB  is  common,  BD  =  BO.  (Cons.) 

But  AC  <  AD.  (497) 

.-.  ZABO  <  ZABD.  (120) 

Q.E.D. 
EXERCISES. 

1.  All  points  whose  projections  upon  a  plane  lie  in  a 
straight  line  are  themselves  in  one  plane:  how  is  this  plane 
defined  ? 

2.  If  AB,  BO,  CD  are  lines,  such  that  ABO,  BOD  are 
right  angles,  and  AB  is  at  right  angles  to  the  plane  BCD, 
then  will  CD  be  at  right  angles  to  the  plane  ABC. 


274  SOLID  GEOMETRY, 


Proposition  2 1 .    Theorem.* 

551.  Between  tiDO  straight  lines  not  in  the  same  plane 
one,  and  only  one,  common  perpendicular  can  le  draiun. 

Hyp,    Let  AB  and   CD  be  the 
given  st.  lines. 

To  prove  there  is  one  _L,  and  no 
more,  to  both  AB  and  CD. 

Proof,  Through  one  of  the  lines, 


B' 


/ 


as  CD,  pass  a  plane  MN  ||  to  the     /      "\""             U 
line  AB.  (514)     ^ 'N 

Through    AB     pass    the    plane 

AB'  _L  to  MN,  and  let  their  intersection  A'B'  meet  CD  at  A'. 
Then  A'B'  is  ||  to  AB.  (517) 

At  A'  erect  A'A  in  the  plane  AB'  J_  to  A'B'. 

.  • .  A'A  is  J_  to  the  plane  MN,  (537) 

and  .  • .  A'A  is  J.  to  the  line  CD.  (487) 

Because  A'A  is  _L  to  A'B',  .• .  it  is  i.  to  AB.  (71) 

.  • .  A'A  is  J_  to  both  the  lines  AB  and  CD. 
If  there  is  any  other  common  _L  to  AB  and  CD,  let  it  be 

HG. 
Because  HC  is  _L  to  AB, 
.'.it  is_Lto  a  line  HL  drawn  ||  to  AB  in  the  plane 

MN,  (71) 

and  .  • .  HG  is  J.  to  the  plane  MN.  (500) 

But  GK  in  the  plane  AB',  _L  to  A'B',  is  J.  to  MN.    (537) 
.  • .  from  the  pt.  G  there  are  two  _Ls  GK,  GH  to  MN. 
But  this  is  impossible.  (502) 

.  • .  HG  is  not  a  common  _L  to  the  lines  AB,  CD. 
.*.  A'A  is  the  only  common  _L  to  AB  and  CD.  q.e.d. 

552.  Cor.  The  perpendicular  A'A  is  the  least  distance 

hetiueen  AB  and  CD  ;  for  any  other  line  GH  >  GK  (497), 

or  its  equal  A'A. 


BOOK  VL-EXERG1SE8.  275 


EXERCISES 


1.  Parallel  lines  intersecting  the  same  plane  make  equal 
angles  with  it.  (492) 

2.  If  a  plane  bisects  a  line  perpendicularly,  every  point 
of  the  plane  is  equally  distant  from  the  extremities  of  the 
line. 

3.  If  three  lines  in  space  are  parallel,  in  how  many  planes 
may  they  lie  when  taken  two  at  a  time  ? 

4.  If  four  lines  in  space  are  parallel,  in  how  many  planes 
may  they  lie  when  taken  two  at  a  time  ? 

5.  How  many  different  planes  may  the  sides  of  a  quad- 
rilateral in  space  contain  when  taken  two  and  two  ? 

6.  If  two  lines  not  in  the  same  plane  are  intersected  by 
the  same  line,  how  many  planes  may  be  determined  by  the 
three  lines  taken  two  and  two  ? 

7.  A  straight  line  makes  equal  angles  with  parallel 
planes. 

8.  The  sum  of  two  adjacent  diedral  angles,  formed  by 
one  plane  meeting  another,  is  equal  to  two  right  diedral 
angles. 

9.  If  two  planes  intersect  each  other,  the  opposite  or 
vertical  diedral  angles  are  equal. 

10.  When  a  plane  intersects  two  parallel  planes,  the 
alternate-interior  diedral  angles  are  equal,  and  the  exterior- 
interior  diedral  angles  are  equal. 

11.  Show  that  two  observations  with  a  spirit-level  are 
sufi&cient  to  determine  if  a  plane  is  horizontal:  and  prove 
that  for  this  purpose  the  two  positions  of  the  level  must 
not  be  parallel. 

12.  To  draw  a  straight  line  perpendicular  to  a  given  plane 
from  a  given  point  outside  of  it. 

13.  To  draw  a  straight  line  perpendicular  to  a  given 
plane  from  a  given  point  in  the  plane. 


^76  SOLID  GEOMETRY, 


POLYEDRAL  ANGLES. 
DEFINITIONS. 

553.  When  three  or  more  planes  meet  in  a  common 
point,  they  are  said  to  form  a  polyedral  angle  at  that  point. 

The  common  point  in  which  the  planes  meet  is  the  vertex 
of  the  angle,  the  intersections  of  the  planes  are  the  edges, 
the  portions  of  the  planes  between  the  edges 
are  t\iQ  faces,  and  the  plane  angles  formed  by 
the  edges  are  i\iQ  face-angles. 

Thus,  the  point  S  is  the  vertex,  the  straight 
lines  SA,  SB,  etc.,  are  the  edges,  the  planes 
SAB,  SBC,  etc.,  are  the  faces,  and  the  angles 
ASB,  BSO,  etc.,  are  the  face-angles  of  the 
polyedral  angle  S  -  ABCD. 

554.  The  edges  of  a  polyedral  angle  may  be  produced 
indefinitely  ;  but  to  represent  the  angle  clearly,  the  edges 
and  faces  are  supposed  to  be  cut  off  by  a  plane,  as  in  the 
figure  above.  The  intersection  of  the  faces  with  this  plane 
forms  a  polygon,  as  ABCD,  which  is  called  the  base  of  the 
polyedral  angle. 

555.  In  a  polyedral  angle,  each  pair  of  adjacent  faces 
forms  a  diedral  angle,  and  each  pair  of  adjacent  edges 
forms  a  face-angle.  There  are  as  many  edges  as  faces,  and 
therefore  as  many  diedral  angles  as  faces. 

556.  The  magnitude  of  a  polyedral  angle  depends  only 
upon  the  7'elatlve  position  of  its  faces,  and  is  independent 
of  their  extent.  Thus,  by  the  face  SAB  is  not  meant  the 
triangle  SAB,  but  the  indefinite  plane  between  the  edges 
SA,  SB  produced  indefinitely. 

557.  Two  polyedral  angles  are  equal,  when  the  face  and 


BOOK   Vl.—POLYEDRAL  ANGLES.  277 

diedral  angles  of  one  are  respectively  equal  to  the  face  and 
diedral  angles  of  the  other,  taken  in  the  same  order, 

558.  A  polyedral  angle  of  three  faces  is  called  a  triedral 
angle;  one  of  four  faces  is  called  a  tetraedral  angle;  etc. 

559.  A  polyedral  angle  is  convex  when  its  base  is  a  con- 
vex polygon.  (141) 

560.  A  triedral  angle  is  called  isosceles  when  it  has  two 
of  its  face-angles  equal ;  when  it  has  all  three  of  its  face- 
angles  equal  it  is  called  equilateral. 

561.  A  triedral  angle  is  called  rectangular,  hi-rect- 
angular,  or  tri-rect angular,  according  as  it  has  one,  two, 
or  three,  right  diedral  angles. 

The  corner  of  a  cube  is  a  tri-rectangular  triedral  angle. 

562.  Two  polyedral  angles  are  symmetrical,  when  the 
face  and  diedral  angles  of  one  are  equal  to  the  face  and 
diedral  angles  of  the  other,  each  to  each,  but  arranged  in 
reverse  order. 

Thus,  the  triedral  angles  S  -  ABC,  S' -  A'B'C  are  sym- 
metrical when  the  face-angles 
ASB,  BSC,  CSA  are  equal  re- 
spectively to  the  face-angles 
A'S'B',  B'S'C,  C'S'A',  and  the 
diedral  angles  SA,  SB,  SC  to  the 
diedral  angles  S'A',  S'B',  S'C. 

When  two  polyedral  angles  are 
symmetrical,  it  is,  in  general,  impossible  to  bring  them  into 
coincidence. 

The  two  hands  are  an  illustration.  The  right  hand  is 
symmetrical  to  the  left  hand,  but  cannot  be  made  to  coin- 
cide with  it.  The  right  glove  will  not  fit  the  left  hand, 
but  is  symmetrical  to  it. 

563.  Opposite  or  vertical  polyedral  angles  are  those  in 
which  the  edges  of  one  are  the  prolongations  of  the  edges 
of  the  other. 


278  SOLID  GEOMETRY, 

Proposition  22.    Theorem. 
564.   Two  opposite  2)olyedral  angles  are  symmetrical. 


B 
Hyp.  Let  S  -  ABC,  S  -  A'B'C  be  two  opp.  triedral  Z  s. 

To  prove  they  are  symmetrical. 

Proof.  Because  the  face  Z  s  ASB  and  A'SB'  are  vertical 
Zs, 

.-.  ZASB=  ZA'SB'  (53) 

Similarly,  Z  BSC  =  Z  B'SC,  etc. 

Also,  because  the  diedral  Z  between  two  planes  is  the 
same  at  every  pt.,  (528) 

.  * .  the  diedral   Z  s  whose  edges  are  SA,  SB,  etc.,  =  re- 
spectively the  diedral  Z  s  whose  edges  are  SA',  SB',  etc. 

But  the  edges  of  S  -  A'B'C  are  arranged  in  the  reverse 
order  from  the  edges  of  S  -  ABO. 

.• .  S  -  ABC  is  symmetrical  to  S  -  A'B'C.         q.e.d. 

EXERCISE. 

Pass  two  parallel  planes,  one  through  each  of  two  straight 
lines  which  do  not  meet  and  are  not  parallel. 

Let  AB,  CD  be  the  lines  :  draw  AE  ||  to  CD,  CF  ||  to  AB.  .*.  plane  AEB  is  li  tg 
plane  CFD. 


BOOK  VL—TBIEDRAL  ANGLES. 


279 


;^ 


Proposition  23.    Theorem. 

565.  The  sum  of  any  two  face-angles  of  a  triedral  angle 
is  greater  than  the  third, 

S 


,V'^C 


Hyp.  Let  S-ABO  be  a  triedral  Z  in  which  /ASC  is 

the  greatest  face  / . 

To  prove  '  Z  ASB  +  Z  BSC  >  Z  ASC. 

Proof,  In  the   plane  ASC  draw  SD, 

making  Z  ASD=:ZASB. 

Draw  AC  cutting  SD  in  D,  take  SB  =  SD,  and  join  AB, 

BC. 
Because  the  Z  s  ASB,  ASD  have 

SA  common,    SB  =  SD,    ZASB  =ZASD,         (Cons.) 
.-.  aASB=aASD,.-.  AB=AD.  (104) 

In  A  ABC,  AB  +  BC>  AC.  (96) 

But  AB  =  AD.  (Proved  above) 

Subtracting,  BC  >  DC.  (Ax.  5) 

Because  in  the  A  s  BSC,  DSC, 

SC  i^  common,  SB  =  SD,  and  BC  >  DC, 

.-.   ZBSC>  ZDSC.  (120) 

But  ZASB=ZASD.  (Cons.) 

Adding,      Z  ASB  +  Z  BSC  >  Z  ASC.  Q.e.  d. 


280 


SOLID   GEOMETRY. 


Proposition  24.    Theorem. 

566.   The  sum  of  the  face-cinglcs  of  any  convex  poly edral 
angle  is  less  tha^ifour  right  angles. 

Hyp.  Let  tlie  convex  »polycdral  /  S 
be  cut  by  a  plane  making  tlie  section 
ABODE  a  convex  polygon. 

To  2^rove  Z  ASB  +  I BSC,  etc., 
<  4  rt.  Z  s. 

Proof  From  any  pt.  0  within  the 
polygon  ABODE  draw  OA,  OB,  00, 
OD,  OE. 

There  are  thus  formed  two  sets  of 
A  s,  one  with  their  common  vertex  at 
S,  and  the  other  with  their  common  vertex  at  0,  and  an 
equal  number  of  each. 

.  • .  the  sum  of  the  Z  s  of  these  two  sets  of  A  s  is  equal.  (97) 

Because  the  sum  of  any  two  face  Z  s  of  a  triedral  Z 
>  the  third,  (565) 


and 


.-.   ZSAE+  ZSAB>ZEAB, 
Z  SBA  +  Z  SBO  >  Z  ABO,  etc. 


Taking  the  sum  of  these  inequalities,  we  find  that  the 
sum  of  the  Z  s  at  the  bases  of  the  A  s  whose  common  ver- 
tex is  S  >  the  sum  of  the  Z  s  at  the  bases  of  the  A  s  whose 
common  vertex  is  0. 

.•.  the  sum  of  the  Zs  at  S  <  tlie  sum  of  the  Zs  at  0. 


the  sum  of  the  Z  s  at  S  <  4  rt.  Z  s. 


(5G) 

Q.E.D. 


BOOK   VL—TRIEDUAL  ANGLES. 


281 


Proposition  25.    Theorem. 

567.  Two  triedral  angles y  which  have  the  three  face- 
angles  of  the  one  equal  respectively  to  the  three  face-a7igles 
of  the  other,  are  either  equal  or  symmetrical. 


B  •■    B'  B' 

Hyp,  In  the  triedral  Zs,  S  and  S',  let  the  face  Zs  ASB, 
BSO,  OSA  =  the  face  Z  s  A'S'B',  B'S'C,  C'S'A'. 

To  prove  S  -  ABC  =  or  symmetrical  to  S'  -  A'B'C. 

Proof,  In  the  edges  SA,  S'A',  etc.,  take  the  six  equal 
distances  SA,  SB,  SO,  S'A',  S'B',  S'C;  and  join  AB,  BO, 
CA,  A'B',  B'O',  O'A'. 

Then,  AS  SAB,  SBO,  SO  A  =  As  S'A'B',  S'BT/,  S'O'A'. 
.-.  A  ABO  =  aA'B'O'.  (108) 

In  the  edges  SA,  S'A',  take  SD  =  S'D'. 

In  the  faces  ASB,  ASO,  and  A'S'B',  A'S'C,  draw  DH, 
DK,  and  D'H',  D'K'  _L  to  AS  and  A'S',  meeting  AB,  AC 
and  A'B',  A'C,  in  H,  Kand  H',  K'.   Join  HKand  H'K'. 

Because  AD  =  A'D',  and  Z  DAH  =  Z  D'A'H', 

.  • .  rt.  A  ADH  =  rt.  A  A'D'H'.  (107) 

.• .  AH  =  A'H',  and  DH  =  D'H'. 

In  the  same  way,  AK  ==:  A'K',  and  DK  =  D'K'. 

.  • .   A  AHK  =  A  A'H'K',  and  HK  =  H'K'.  (104) 

.  • .   A  HDK  =  A  H'D'K',  and  Z  HDK  =  Z  H'D'K'.  (108) 
.  • .  diedral  Z  SA  =  diedral  Z  S'A'.  (536) 

Similarly,  diedral   Zs  SB,  SO  =  diedral  ZsS'B',  S'C. 

Now  if  the  ^ual  Z  s  are  arranged  in  the  same  order,  as  in 
the  first  two  figures,  the  two  triedral  Z  s  are  equal.        (557) 

P>ut  if  the  equal  Z  s  are  in  reverse  order,  as  in  the  first 
and  third  figures,  the  triedral  Z  s  are  symmetrical.      (5G2) 

Q.E.D, 


282  SOLID  GEOMETRY. 


EXERCISES. 

Theorems. 

1.  If  a  straight  line  is  parallel  to  a  plane,  any  plane  per- 
pendicular to  the  line  is  perpendicular  to  the  plane. 

See  (503)  and  (540). 

2.  If  a  line  is  parallel  to  each  of  two  intersecting  planes, 
it  is  parallel  to  their  intersection. 

3.  If  a  line  is  perpendicular  to  a  plane,  every  plane  paral- 
lel to  that  line  is  also  perpendicular  to  the  plane. 

4.  If  a  line  is  parallel  to  each  of  two  planes,  the  intersec- 
tions which  any  plane  passing  through  it  makes  with  the 
planes  are  parallel. 

5.  If  a  straight  line  and  a  plane  are  perpendicular  to  the 
same  straight  line,  they  are  parallel. 

6.  If  two  straight  lines  are  parallel,  they  are  parallel  to 
the  common  intersection  of  any  two  planes  passing  through 
them. 

7.  If  the  intersections  of  several  planes  are  parallel,  the 
normals  drawn  to  them  from  any  point  are  in  one  plane. 

8.  If  a  plane  is  passed  through  one  of  the  diagonals  of  a 
parallelogram,  the  perpendiculars  to  it  from  the  extremities 
of  the  other  diagonal  are  equal. 

9.  Prove  that  the  sides  of  ah  isosceles  triangle  are  equally 
inclined  to  any  plane  through  the  base. 

10.  From  a  point  P,  PA  is  drawn  perpendicular  to  a  given 
plane,  and  from  A,  AB  is  drawn  perpendicular  to  a  line 
in  that  plane:  prove  that  PB  is  also  perpendicular  to  that 
line. 

11.  If  the  projections  of  any  line  upon  two  intersecting 
planes  are  each  of  them  straight  lines,  prove  that  the  line 
itself  is  a  straight  line. 

See  (495). 

12.  If  a  plane  is  passed  through  the  middle  point  of  the 
common  perpendicular  to  two  straight  lines  in  space,  and 
parallel  to  both  of  these  lines,  prove  that  the  plane  bisects 


BOOK   VI.— EXERCISES.     THEOREMS.  283 

every  straight  line  which  joins  a  point  of  one  of  these  lines 
to  a  point  of  the  other. 

See  (526). 

13.  The  projection  of  a  straight  line  on  a  plane  is  a 
straight  line  which  is  either  parallel  to  the  first  straight 
line  or  meets  it  in  the  point  where  it  cuts  the  plane,  ac- 
cording as  the  first  line  is  parallel  to  the  plane  or  not. 

14.  From  a  point  P,  PA  and  PB  are  drawn  perpendic- 
ular to  two  planes  which  intersect  in  CD,  meeting  them  in 
A  and  B;  from  A,  AE  is  drawn  perpendicular  to  CD:  prove 
that  BE  is  also  perpendicular  to  CD. 

15.  Two  planes  which  are  not  parallel  are  cut  by  two 
parallel  planes:  prove  that  the  intersections  of  the  first  two 
with  the  last  two  contain  equal  angles. 

See  (519),  (525). 

16.  If  two  parallel  planes  be  cut  by  three  other  planes 
which  have  a  point,  but  no  line  common  to  all  three,  and 
no  two  of  which  are  parallel,  the  triangles  formed  by  the 
intersections  of  the  parallel  planes  with  the  other  three 
planes  are  similar  to  each  other. 

See  (519),  (525). 

17.  The  projections  of  parallel  straight  lines  on  any 
plane  are  themselves  parallel. 

See  (509),  (525),  (519). 

18.  Two  straight  lines  which  intersect  are  inclined  to 
each  other  at  an  angle  equal  to  two-thirds  of  a  riglit  angle, 
and  to  a  given  plane,  each,  at  an  angle  equal  to  half  a  right 
angle.  Prove  that  their  projections  on  this  plane  are  at 
right  angles  to  each  other. 

19.  In  any  triedral  angle,  the  planes  bisecting  the  three 
diedral  angles,  all  intersect  in  the  same  straight  line. 

See  (549). 

20.  In  any  triedral  angle,  the  planes  which  bisect  the  three 
face-angles,  and  are  perpendicular  to  these  faces  respect- 
ively, all  intersect  in  the  same  straight  line. 

From  the  vertex  S,  take  equal  distances  SA,  SB,  SC,  on  the  three  edges;  the 
intersections  of  the  three  ±  bisecting  planes  with  the  plane  of  ABC  are  ±  bisect- 
ing lines  of  the  sides  of  ABC,  ^nd  h^ve  a  common  intersection  (168);  .• .  etc. 


284  SOLID  GEOMETRY. 

21.  In  any  triedral  angle,  the  three  planes  which  pass 
through  the  edges,  perpendicular  to  the  opposite  faces 
respectively,  all  intersect  in  the  same  straight  line. 

See  (546). 

23.  ABCD  is  a  face  and  AE  a  diagonal  of  a  cube,  BG  is 
drawn  perpendicular  to  AE,  and  DG  is  joined :  prove  that 
DG  is  perpendicular  to  AE. 

23.  If  two  face-angles  of  a  triedral  angle  are  equal,  the 
diedral  angles  opposite  them  are  also  equal. 

24.  An  isosceles  triedral  angle  and  its  symmetrical  tri- 
edral angle  are  equal. 

25.  If  BAG,  CAD,  DAB  be  the  three  face-angles  of  a 
triedral  angle,  prove  that  the  angle  between  AD  and  the 
straight  line  bisecting  the  angle  BAG  is  less  than  half  the 
sum  of  the  angles  BAD,  CAD. 

See  (565). 

26.  A  triedral  angle  is  contained  by  the  three  face-angles 
BOG,  CO  A,  AOB;  if  BOG,  COA  are  together  equal  to  two 
right  angles,  prove  that  CO  is  perpendicular  to  the  line 
which  bisects  the  angle  AOB. 

Loci. 

27.  Find  the  locus  of  points  which  are  equally  distant 
from  three  given  points  not  in  the  same  straight  line. 

28.  Find  the  locus  of  points  which  are  equally  distant 
from  two  given  intersecting  straight  lines. 

29.  Find  the  locus  of  points  which  are  equally  distant 
from  two  given  parallel  planes;  or  whose  distances  from 
the  parallel  planes  are  in  a  given  ratio. 

30.  Find  the  locus  of  points  which  are  equally  distant 
from  three  given  planes. 

See  (549). 

31.  Find  the  locus  of  points  which  are  equally  distant 
from  the  three  edges  of  a  triedral  angle. 

32.  Find  the  locus  of  points  which  are  equally  distant 
from  the  three  faces  of  a  triedral  angle. 

33.  Find  the  locus  of  pointG  whicli  are  equally  distant 


I 


BOOK  Vl—EXERCISES     PROBLEMS.  "^St^ 


from  two  given  planes,  and  at  the  same  time  equally 
distant  from  two  given  points. 

34»  Find  the  locus  of  a  point  such  that  the  sum  of  its 
distances  from  two  given  planes  is  equal  to  a  given  straight 
line. 

35.  Find  the  locus  of  a  point  such  that  the  sum  of  its 
distances  from  three  given  planes  is  equal  to  a  given  straight 
line. 

Problems. 

3G.  Pass  a  plane  perpendicular  to  a  given  straight  line 
through  a  given  point  not  in  that  line. 

37.  Pass  a  plane  through  a  given  straight  line,  perpen- 
dicular to  a  given  plane. 

See  (543). 

38.  Pass  a  plane  through  a  given  point  parallel  to  a  given 
plane. 

See  (522). 

39.  Find  the  point  in  a  given  straight  line  which  is 
equally  distant  from  two  given  points  not  in  the  same  plane 
with  the  given  line. 

40.  Draw  a  straight  line  through  a  given  point  in  space, 
so  that  it  shall  cut  two  given  straight  lines  not  in  the  same 
plane. 

41.  Draw  a  straight  line  through  a  given  point  in  a  given 
plane,  so  that  it  shall  be  perpendicular  to  a  given  line  in 
space. 

42.  Two  given  straight  lines  do  not  intersect  and  are 
not  parallel :  find  a  plane  on  which  their  projections  will 
be  parallel. 

43.  Divide  a  straight  line  similarly  to  a  given  divided 
straight  line  lying  in  a  different  plane. 

Let  ACDB  be  the  divided  line,  NMLF  the  other  line  ;  draw  EHKG II  to  AB,  and 
CH,  DK,  BG  II  to  AE  ;  also  KL,  HM.  EN  II  to  GF:  .• .  (525),  (526). 

44.  Given  three  straight  lines  meeting  at  a  point:  draw 
through  the  given  point  a  straight  line  equally  inclined  to 
the  three. 


Book  VII. 

POLYEDKONS. 


Definitioi^s. 


568.  K  polyedron  is  a  solid  bounded  by  planes.  The 
portions  of  the  bounding  planes,  limited  by  their  mutual 
intersections,  are  the  faces  of  the  polyedron;  tlie  intersec- 
tions of  the  faces  are  the  edges,  and  the  intersections  of  the 
edges  are  the  vert  ices,  of  the  polyedron. 

A  diagonal  of  a  polyedron  is  a  straight  line  joining  two 
vertices  not  in  the  same  face. 

569.  A  polyedron  of  four  faces  is  called  a  tetraedron; 
one  of  six  faces,  a  hexaedron;  one  of  eight  faces,  an  octa- 
edron;  one  of  twelve,  a  dodecaedron;  one  of  twenty,  an 
tcosaedron. 

5  70.  A  polyedron  is  convex  when  the  section  made  by 
any  plane  intersecting  it  is  a  convex  polygon. 

Only  convex  polyedrons  are  treated  of  in  this  work. 

571.  The  volume  of  a  solid  is  its  numerical  measure 
(229),  referred  to  some  other  solid  called  the  unit  of  vol- 
ume. 

572.  Two  solids  are  equivalent  when  they  have  equal 
volumes. 

573.  Two  polygons  are  said  to  be  parallel  when  their 
sides  are  respectively  parallel. 

286 


BOOK  VII.—P0LYEDR0N8.    DEFINITIONS,        287 


Prisms  and  Parallelopipeds. 

5  74.  A  prism  is  a  polyedron  two  of  whose  faces  are 
equal  and  parallel  polygons,  and  the  other 
faces  are  parallelograms. 

The  equal  and  parallel  polygons  are 
called  the  bases  of  the  prism;  the  par- 
allelograms are  the  late7'al  faces;  the  lat- 
eral faces  taken  together  form  the  lateral 
or  convex  surface;  and  the  intersections 
of  the  lateral  faces  are  the  lateral  edges. 

The  lateral  edges  are  parallel  and  equal  (511)  and  (520). 
The  area  of  the  lateral  surface  is  called  the  lateral  area. 

575.  The  a//iY?<^e  of  a  prism  is  the  perpendicular  dis- 
tance between  its  bases. 

576.  A  prism  is  said  to  be  triangular,  quadrangular, 
hexagonal,  etc.,  according  as  its  bases  are  triangles,  quadri- 
laterals, hexagons,  etc. 

577.  K  rigid  prism  is  one  whose  lateral 
edges  are  perpendicular  to  its  bases. 

A  regular  prism  is  a  right  prism  whose 
bases  are  regular  polygons :  therefore  its  lat- 
eral faces  are  equal  rectangles. 

An  oblique  prism  is  one  whose  lateral  edges 
are  not  perpendicular  to  its  bases. 

578.  A  right  section  of  a  prism  is  a  sec- 
tion by  a  plane  perpendicular  to  its  lateral 
edges. 

579.  A  truncated  prism  is  the  part  of  a 
prism  included  between  the  base  and  an  ob- 
lique section  made  by  a  plane  cutting  all  the 
lateral  edges. 

580.  A  parallelopi2)ed  is  a  prism  whose 
bases  are  parallelograms :  hence  all  the  faces 
are  parallelograms. 

581.  K  right  ptarallelopiped  \s  one  whose 
lateral  edges  are  perpendicular  to  the  bases: 
hence  the  lateral  faces  are  rectangles. 


"^^ij 


288 


SOLID  GEOMETRY. 


582.  A  rectangular  parallehpvpcd  is  one 
whose  faces  are  all  rectangles. 

The  three  edges  of  a  parallelepiped  which 
meet  at  a  vertex  are  its  dimensions. 


o 


583.  A  ciihe  is  a  rectangular  parallelepiped 
whose  six  faces  are  all  squares. 

584.  The   cube   whose   edge   is  the  unit  of 
length  is  taken  as  the  unit  of  volume.  (^'^1) 

Eem.  In  a  right  parallelepiped  the  adjacent 
lateral  faces  may  make  any  angle  with  each  other,  while 
in  a  rectangular  parallelepiped  the  faces  are  perpendicular 
to  each  other. 


Proposition   1 .    Theorem. 

585.  The  sections  of  the  lateral  faces  of  a  prism  made 
hy  parallel  pkmes  are  equal  polygons.  ^^^ 

Hyp.  Let  the  prism  MN  be  cut  by 
the  II  planes  AD,  A'D'. 

To  prove  ABODE  =  A'B'C'D'E'. 
Proof  Because  the  intersections  of 
two  II  planes  by  a  third  plane  are  par- 
allel, (519) 
.*.  AB,  BC,  CD,  etc.,  are  ||  respec- 
tively to  A'B',  B'C,  CD',  etc. 

.-.   ZABC:=ZA'B'C',  ZBCD  =  ZB'C'D'.   (525) 
Because   ||  s  included  between   ||  s  are  equal,  (1^1) 

.  • .  AB  =  A'B',  BC  =  B'C,  etc. 
.• .  the  polygons  ABCDE,  A'B'C'D'E'  are  mutually  equi- 
lateral and  mutually  equiangular. 

.• .  ABCDE  =  A'B'C'D'E'.  (14G) 

Q.E.D. 

586.  Cor.  Any  section  of  a  prism  hy  a  plane  parallel 
to  the  base  is  equal  to  the  base;  and  all  right  sections  are 
equal 


BOOK  VIL— EQUAL  PRISMS. 


289 


(Hyp.) 

(567) 


Proposition  2.    Theorem. 

587.  If  three  faces  including  a  triedral  angle  of  a  prism 
are  equal  respectively  to  three  faces  including  a  tried^rd 
angle  of  a  second  pris?n,  and  siniilaidy  placed,  the  two  prisms 
are  equal. 

Hyp.  Let  the  faces  AD, 
AG,  BH,  of  the  triedral 
Z  B,  be  =  respectively  to 
thefacesA'D',A'a',B'H' 
of  the  triedral  Z  B'. 

To  prove 
prism  AI  =  prism  A'l'. 

Proof,  Since  the  face 
ZsABO,ABG,CBG  =  the 
face  ZsA'B'C,  A'B'G',  C'B'G',  respectively, 

.  • .  triedral  Z  B  =  triedral  Z  B'. 

.  • .  the  prism  AI  may  be  applied  to  the  prism  A'l'  so 
tliat  the  base  AD  shall  coincide  with  A'D',  the  face  AG 
with  A'G',  and  the  face  BH  with  B'H',  the  pts.  D,  E  fall- 
ing on  D',  E'. 

Since  the  pts.  F,  G,  H  coincide  with  F',  G',  H', 
.  • .  the  planes  of  the  upper  bases  will  coincide.  (484) 

Since  the  lateral  edges  of  the  prisms  are  equal  and  par- 
allel, (574) 
.-.  the  edges  DI,  EK  will  coincide  with  D'l',  E'K',  re- 
spectively, and  the  pts.  I,  K  with  the  pts.  V,  K'. 

.*.  the  prisms  coincide  throughout,  and  .*.  are  equal. 

Q.E.D. 

588.  CoK.  1.  Two  right  prisms  which  have  equal  bases 
and  equal  altitifdes  are  equal, 

589.  CoK.  2.  Two  truncated  prisms  are  eqiial  if  three 
faces  including  a  triedral  angle  of  the  one  are  equal  re- 
spectively to  three  faces  including  a  triedral  angle  of  the 
other y  and  similarly  placed. 


290  SOLID  GEOMETRY. 

Proposition  3.    Theorem. 

590.  An  oUique  prism  is  equivalent  to  a  right  prism 
havi7igfor  its  lase  a  right  section  of  the  oblique  prism  and 
for  its  altitude  u  lateral  edge  of  the  oblique  prism. 


B         C 

Hyp,  Let  ABCDE-I  be  an  oblique  prism,  and  A'D'  art. 
section  of  it. 

Produce  AF  to  F',  making  A'F'  =  AF,  and  through  F' 

pass  the  plane  F'l'  i.  to  AF',  cutting  all  the  edges  BG, 

CH,  etc.,   produced,  in  the  pts.  G',  H',  etc.,  and  forming 

the  rt.  section  F'l'  ||  to  A'D'.  (51G) 

To  prove  prism  AI  =  prism  A'l'. 

Proof  In  the  truncated  prisms  AD'  and  FI', 

ABODE  =  FGHIK.  (574) 

Because       AF  =  A'F',  and  BG  =  B'G', 
.• .  AA'  =  FF',  and  BB'  =  GG'. 
Because  AG  and  A'G'  are  CJs,  (574) 

.• .  AB  is  =  and  ||  to  FG,  and  A'B'  is  =  and  ||  to  F'G'. 

.-.  quadl.  AB'  =  quadl.  FG', 
being  mutually  equilatei'al  and  equiangular  (146). 
Similarly,  BC  =  GH'. 

.  •.  the  truncated  prisms  AD'  and  FI'  are  equal.       (589) 
Taking  AD'  from  the  whole  solid,  the  right  prism  A'l' 
remains,  and  taking  FI'  from  the  same  solid,  the  oblique 
prism  AI  remains. 

.*.  prism  AI  =  prism  A'l'.  q.e.d. 


BOOK   VIL—LATEBAL  AREA   OF  A  PRISM.        291 


Proposition  4.    Theorem. 

591.   The  lateral  area  of  a  ^jrism  is  equal  to  the  product 
of  the  ])erimeter  of  a  right  section  by  a  lateral  edge. 


B 

Hyp,  Let  FGHIK   be  a  rt.  section,  and  AA'  a  lateral 
edge  of  the  prism  AD'. 
To  /jroz^e  lateral  area  of 

AD'  =  AA'  (FG  +  GH  +  etc.). 

Proof  Since   the  rt.   section   FI  is   J.  to    the    lateral 

edges,  (578) 

.• .  FG,  GH,  etc.,  are  _L  to  AA',  BB',  etc.,       (487) 

and        .* .  FG,  GH,  etc.,  are  the  altitudes  of  the  /Z7s  AB', 

BC,  etc.,  which  form  the  lateral  area  of  the  prism. 

.-.  area  AA'B'B  =  A  A'  X  FG,  (363) 

and  area  BB'C'C  =  BB'  X  GH,  etc. 

Because  AA'  =  BB'  =  etc.,  (574),  and  FG  +  GH  + 
etc.,  =  the  perimeter  of  the  rt.  section,  .'.  adding  the  above 
areas,  we  have 

lateral  ai-ea  AD'  =  AA'  X  FG  +  AA'  X  GH  +  etc. 

=  AA'(FG+GH+HI+etc.). 

Q.E.D. 

592.  Cor.  The  lateral  area  of  a  right  prism  is  equal  to 
the  product  of  the  perimeter  of  its  base  by  its  altitude. 


292  SOLID  GEOMETHY. 


'  Proposition  5.    Theorem. 

693.  Tlie  ojjjjosite  faces  of  a  paraUelopiped  are  equal 
and  parallel. 


Eic aH 


B  C 

Hyp.  Let  ABCD-G  be  a  paraUelopiped. 
To  prove  face  AF  =  and  ||  to  DG. 

Proof  Since  AC  is  a  z=7,  (580) 

.-.  AB  =  and||toDC. 
Since  BG  is  a  OJ,  (580) 

.-.  BF  =  and  II  to  CG. 
Since  BA  and  BF  are  ||  respectively  to  CD  and  CG, 
.• .  Z  ABF  =  Z  DCG,  and  plane  AF  is  ||  to  plane  DG.(525) 
.-.  ^  AF  =^DG, 
having  two  sides  and  the  included  Z  equal,  each  to  each  (135). 
.'.  face  AF  =  and  ||  to  face  DG. 
It  may  be  proved  in  the  same  way  that 

face  AH  =  and  ||  to  face  BG. 
Also,   face  AC  =  and  ||  to  face  EG,  by  definition.     (574) 

Q.E.D. 

594.  ScH.  Since  the  opposite  faces  of  a  parallelepiped 
are  equal  and  parallel  parallelograms,  any  face  may  be  taken 
for  the  base. 

695.  Cor.  Hie  twelve  edges  of  a  paraUelopiped  may  he 
divided  into  three  sets,  each  set  having  four  equal  and  par- 
allel lines. 


BOOK   VIl.—PARALLELOPIPEDa, 


293 


Proposition  6.    Theorem. 

596.  Tlie  four  diago7ials  of  a  parallelopiped  bisect  one 
another. 

Hyp.  Let  ABCD-G  be  a  paral- 
lelopiped. 

To  prove  that  the  four  diagonals 
r   AG,  BH,  CE,   DF  bisect  one  an- 
other. A< 

Proof.  Through  the  opp.  and  || 
edges  AE,  CG  pass  a  plane.  Join 
AC,  EG. 

Because  AE  and  CG  are  =  and  H,  (574) 


.  • .  the  figure  ACGE  is  a  E7. 

.*.  its  diagonals  AG,  EC  bisect  each  other  in  the  pt.  0.  (134) 
In  the  same  way  it  may  be  shown  that  AG,  BH,  and  AG, 
DF,  bisect  each  other  at  0. 

.  • .  the  four  diagonals  bisect  one  another  at  the  pt.  0. 

Q.E.D. 

597.  Cor.  1.  The  diagonals  of  a  rectangular  parallelo- 
piped are  equal. 

598.  Cor.  2.  The  sq^iare  of  a  diagonal  of  a  rectangular 
parallelopiped  is  equal  to  the  sum  of  the  squares  of  the 
three  edges  meeting  at  any  vertex. 

For,  if  AG  is  a  rectangular  parallelopiped,  the  rt.  A  s 
ACG,  ABC  give 

AG'  =  AC"  +  CG'  =  AB'  +  BC'  +  BF'. 

599.  ScH.  The  point  0  through  which  all  four  diagonals 
pass  is  called  the  centre  of  the  parallelopiped. 


294 


80LID  QEOMETBT. 


Proposition.  7.    Theorem. 


600.  The  plane  passed  through  kvo  diagonally  opposite 
edges  of  a  parallelopiped  divides  it  into  two  equivalent  tri- 
angular pris7ns. 

Hyp,  Let  the  parallelopiped 
ABCD-G  be  divided  by  the  plane 
AEGO,  passing  through  the  opp. 
edges  AE  and  CG. 

To  prove  that  the  triangular 
prisms  ABC-G  and  ADC-G  are 
equivalent. 

Proof,  Take  a  rt.  section 
KLMO  of  the  parallelopiped,  in- 
tersecting the  plane  AEGO  in  the  line  KM. 

Because  the  faces  AF  and  AH  are  ||  respectively  to  DG 
and  BG,  (593) 


1 

LJ 

^1 

AT"' 

A 

/ 

\ 

i 

'^  - 

-4/ 

/m 

KL  is  II  to  OM,  and  KO  is  ||  to  LM. 


(519) 


KLMO  is  a  O. 


(124) 


.'.  the  intersection  KM  of  the  plane  AEGO  with  the  rt. 
section  is  the  diagonal  of  the  LJ  KLMO,  and  divides  it 
into  two  equal  As  KLM,  KOM.  (130) 

The  oblique  prism  ABO-E  is  equivalent  to  a  rt.  prism 
whose  base  is  KLM  and  altitude  AE,  and  the  oblique  prism 
ADO-E  is  equivalent  to  a  rt.  prism  whose  base  is  KOM  and 
alt.  AE.  (590) 

But  since  these  two  rt.  prisms  have  equal  bases  KLM 
and  KOM,  and  the  same  altitude,  they  are  equal.         (588) 


prism  ABC-G  is  equivalent  to  prism  ADC-G.    q..e.d. 


BOOK   VII.—PARALLEL0PIPED8, 


295 


Proposition  8.    Theorem. 

601.  Two  rectangular    parallelopipeds    having    equal 
bases  are  to  each  other  as  their  altitudes. 

Hyp.  Let  AB  and  CD  be  the 
altitudes  of  two  rectangular  par- 
allelopipeds,  P  and   Q,   having     B 
equal  bases. 

rojt?roveP:Q  =  AB:CD. 

Proof,  Suppose  the  altitudes 
AB  and  CD  have  a  common 
measure,  which  is  contained  m 
times  in  AB  and  n  times  in  CD. 


\ 


V 
V 


Then 


AB  :  CD  =  m  :  n. 


Divide  AB  and  CD  by  this  common  measure. 

Through  the  several  pts.  of  division  of  AB  and  CD  pass 
planes  J_  to  these  lines. 

The  parallelopiped  P  will  be  divided  into  m,  and  Q  into 
n,  parallelopipeds,  all  equal  to  one  another.  (588) 


and 


'.  P  :  Q  =  m  :  7^; 
.  P:Q  =  AB:OD. 


Q.E.D. 


I 


The  case  in  which  the  altitudes  AB  and  CD  are  incom- 
mensurable, is  included  in  this  demonstration  by  the  rea- 
soning in  (232) ;  or  the  proof  may  be  extended  as  in  (234), 
(208),  and  (356). 

602.  ScH.  This  theorem  may  also  be  expressed  as  fol- 
lows : 

Tivo  rectangular  parallelopipeds  tuhich  have  two  dimen- 
sions in  common  are  to  each  other  as  their  third  dimen- 
sions. 


296 


SOLID  GEOMETRY. 


Proposition  9.    Theorem. 

603.  Two  rectangular  imrallelo'pijjeds  having  equal  ah 
titades  are  to  each  other  as  their  bases. 

Hgp.  Let  the  rectangles  ab  and  p  q 

a^y  be  the  bases  of  two  rectangular 
parallelopipeds,  P  and  Q,  having 
the  common  altitude  c. 


m  F       ab 

To  prove       ^  =  -^-^,. 

Proof,  Construct  a  third  paral- 
lelepiped R  whose  dimensions  are 
a,  y,  and  c. 

Then,  because  P  and  R  liave  the 
two  dimensions  a  and  c  in  com- 
mon, (Cons.) 
P_^ 

'''n~b'' 


i\ 

v^ 

E 


(602) 


(602) 


Q.E.D. 


And  because  R  and  Q  have  the  two  dimensions  ¥  and  c 
in  common,  (Cons.) 

R_^ 

Multiplying  these  equalities,  we  have 

P_  ab^ 
Q  ""  a'b' ' 

604.  ScH.  This  theorem  may  also  be  expressed  as  fol- 
lows: 

Two  rectangular  parallelopipeds  having  one  dimension 
in  common  are  to  each  other  as  the  products  of  the  other 
two  dimensions, 

EXERCISE. 

Eind  the  length  of  the  diagonal  of  a  rectangular  paral- 
lelepiped whose  edges  are  1,  4,  and  8. 


BOOK  VIL^PABALLEL0PIPED8. 


297 


Proposition  lO.    Theorem. 

605.  Any  two  rectmigular  paraUelojnpeds  are  to  each 
other  as  the  products  of  their  three  dimensions. 


Hyp,  Let  P  and  Q  be  two  rect- 
angular   parallelopipeds     whose 
dimensions  are  a,  h,  c,  and  a\  ¥y  c',     ^ 
respectively. 

^  P        ale 

Proof.  Make  a  third  rectangu- 
lar parallelepiped  E  whose  dimen- 
sions are  «',  h\  and  c. 

Then,  because  P  and  K  have 
the  dimension  c  in  common^ 


L 


\ 


Of 


R  ~  a'h'' 


(604) 


And  because  R  and  Q  have  the  two  dimensions  «',  h'  in 
common, 

R_  c 


(602) 


Multiplying  these  equalities,  we  have 


P  _    ahc 
Q  ~  a/b'c' 


Q.E.D. 


EXERCISE. 


Find  the  ratio  of  two  rectangular  parallelopipeds  whose 
dimensions  are  4,  7,  9,  and  6,  14,  15,  respectively. 


298 


SOLID  GEOMETRY. 


Proposition  1  1 .    Theorem. 

606.   The  volume  of  a  rectangular  parallelopiped  is  equal 
to  the  product  of  its  three  dime^isions. 

Hyp.  Let  P  be  the  rectangular 
parallelopiped,  a,  h,  and  c  its  dimen- 
sions, and  let  Q  be  the  cube  whose 
edge  is  the  linear  unit.    Q  then  is  the 


lit  of  volume. 

(584) 

To  prove 

V=ahc, 

Proof 

P      ax  b  X  c 
Q~1X IX 1 

But  since 

Qis 

the  unit  of  volume, 

A 


Q 


Q 


=  abc. 


=  the  volume  of  P. 


(605) 


(571) 


Q.E.D. 


.  • .  volume  of  P  =  abc. 

607.  ScH.  The  statement  of  this  theorem  is  an  abbre- 
viation of  the  following : 

TJie  number  of  units  of  volume  in  a  rectangular  parallel- 
opiped  is  equal  to  the  product  of  the  numbers  lohich  measure 
the  linear  units  in  its  three  dimensions. 

Compare  (361). 

When  the  three  dimensions  of  a  rectangular  parallelopiped 
are  each  exactly  divisible  by  the  linear 
unit,  the  truth  of  the  theorem  may  be 
shown  by  dividing  the  solid  into  cubes, 
each  equal  to  the  unit  of  volume. 
Thus,  if  AB  contain  the  linear  unit  3 
times,  AC,  4  times,  and  AD,  5  times, 
these  edges  may  be  divided  respectively 
into  3,  4,  and  5  equal  parts,  and  then 
planes  passed  through  the  several  points  ^  ^ 

of   division   at  right   angles   to   these 
edges  will  divide  the  solid  into  cubes  each  equal  to  the  unit 


\D 


BOOK  VIL—PARALLELOPIPEDS.  299 

of  volume.     Hence  the  whole  solid  contains  3  X  4  X  5,  or 
60  cubes,  each  equal  to  the  unit  of  volume. 

608.  Cor.  1.  Since  a  x  bi^  the  area  of  the  base,  and  c 
is  the  altitude,  of  the  parallelepiped  P ;  therefore  the  above 
result  may  be  expressed  in  the  form  : 

The  volume  of  a  rectangular"  pai^allelopiped  is  equal  to 
the  product  of  its  hase  and  altitude, 

609.  Cor.  2.  The  volume  of  a  cube  is  the  tliird power  of 
its  edge,  being  the  product  of  three  equal  factors ;  if  the 
edge  is  1,  the  volume  is  1  X  1  X  1  =  1;  if  the  edge  isrt,the 
volume  \^  a  X  a  X  a  =^  a^.  Hence  it  is  that  in  arithmetic 
and  algebra  the  ^"^cube"  of  a  number  is  the  name  given  to 
the  "  third  power  '^  of  a  number. 

EXERCISES. 

1.  Find  the  surface,  and  also  the  volume,  of  a  rectangular 
parallelepiped  whose  edges  are  4,  7,  and  9  feet. 

2.  Find  the  surface  of  a  rectangular  parallelepiped  whose 
base  is  8  by  12  feet  and  whose  volume  is  384  cubic  feet. 

3.  Find  the  volume  of  a  rectangular  parallelepiped  whose 
surface  is  208  and  whose  base  is  4  by  6. 

4.  Find  the  length  of  the  diagonal  of  a  rectangular 
parallelepiped  whose  edges  are  3,  4,  and  5. 

5.  Find  the  ratio  of  two  rectangular  parallelepipeds 
whose  dimensions  are  1,  4,  8,  and  3,  4,  5,  respectively. 

6.  Find  the  surface  of  a  rectangular  parallelepiped  whose 
base  is  7  by  9  feet  and  whose  volume  is  315  cubic  feet. 

7.  Find  the  volume  of  a  rectangular  parallelepiped  whose 
surface  is  416  and  whose  base  is  4  by  12. 

8.  A  man  wishes  to  make  a  cubical  cistern  whose  con- 
tents are  186624  cubic  inches:  how  many  feet  of  inch 
boards  will  line  it  ? 

9.  Find  the  side  of  a  cube  which  contains  as  much  as  a 
rectangular  parallelepiped  20  feet  long,  10  feet  wide^  and  6 
feet  high. 


300 


SOLID   GEOMETRY. 


Proposition  1 2.    Theorem. 

610.  Tlie  volume  of  any  parallelopiped  is  equal  to  the 
product  of  its  base  and  altitude. 

Hyp.  Let  B'O  be  the    d',^ ?:..__  h;_ q' 

altitude  of  the   oblique 
parallelopiped  AC. 

To  prove  that 
vol.AC'=ABODxB'0. 


\ 


H' 


:j=' 


Proof.  Produce  the 
edges  AB,  DC,  A'B', 
D'C,  and  take 

EF  =  AB. 


C  M 


The  rt.  parallelopiped  EG',  formed  by  the  sections 
EE'H'H,  FF'G'G  J_  to  the  produced  edge  EF  is  equivalent 
to  AC.  (590) 

Now  produce  the  edges  HE,  GF,  G'F',  H'E',  and  take 
KL  =  HE. 

The  parallelopiped  KNN'K'-L  formed  by  the  sections 
LMM'L',  KNN'K'  J_  to  the  produced  edge  KL  is  equiva- 
lent to  EG',  and  .  • .  to  AC.  (590) 

.*.  given  parallelopiped  AC  and  the  last,  KM',  are 
equivalent. 

But,  since  the  rt.  sections  LM',  KN',  are  rectangles, 

.  • .  KM'  is  a  rectangular  parallelopiped.  (582) 

Also,  since 

area  ABCD  =  area  EFGH  =  area  KLMN,  (363)  (360), 
and  the  three  solids  have  the  same  altitude  B'O,  (523) 

.-.the  volume  KM'  =  KLMN  X  B'O.  (608) 

.-.  the  volume  AC  =  ABCD   X  B'O.         Q.e.d. 


BOOK  VII.- VOLUME  OF  A  PEISM. 


301 


Proposition  13.    Theorem. 

611.  Tlie  volume  of  a  tria7igular  prism  is  equal  to  the 
product  of  its  base  and  altitude,  ,       _       ^, 

Hyp,  Let  H  denote  the  altitude  of 
the  triangular  prism  ABC-B'. 

Toprove  vol.  ABC-B'  =  ABC  X  H. 

Proof  Complete  the  parallelopiped 
ABCD-D',  having  its  edges  =  and  |1  to 
AB,  BC,  BB'. 


vol.  ABC-B'  =  i  vol.  ABCD-B', 

area  ABC  =  ^  area  ABCD. 

vol.  ABCD-D'  =  ABCD  X  H. 

.-.  vol.  ABC-B'  =  J  ABCD  X  H, 
r=  ABC  X  H. 


Then,  vol.  ABC-B'  =  i  vol.  ABCD-B',  (600) 

and  area  ABC  =  1  area  ABCD.  (130) 

But  vol.  ABCD-D'  =  ABCD  X  H.  (610) 

Q.E.D. 

612.  Cor.  1.  The  volume  of  any  prism  is  equal  to  the 
product  of  its  base  and  altitude. 

For,  any  prism  may  be  divided  into 
triangular  prisms  by  passing  planes 
through  a  lateral  edge  AA'  and  the  cor- 
responding diagonals  of  the  base.  Then 
the  volume  of  the  given  prism  is  the  sum 
of  the  volumes  of  the  triangular  prisms, 
that  is,  the  sum  of  their  bases,  or  the 
base  of  the  given  prism,  multiplied  by 
the  common  altitude. 

613.  Cor.  2.  Two  prisms  are  to  each  other  as  the  prod- 
ucts of  their ^  bases  and  altitudes;  two  jt?n^*?m.<?  having 
equivalent  bases  are  to  each  other  as  their  altitudes  ;  ttvo 
prisms  of  the  same  altitude  are  to  each  other  as  their  bases  ; 
two  prisms  having  equivalent  bases  and  equal  altitudes  are 
equivalent. 


302 


SOLID  GEOMETRY. 


Pyramids. 


DEFINITIONS. 


614.  A  pyramid  is  a  polyedron  bounded  by  a  polygon, 
and  by  triangles  meeting  at  a  common  point,  the  sides  of  the 
polygon  being  the  bases  of  the  triangles. 

The  polygon  ABODE  is  called  the 
base  of  the  pyramid;  the  point  S,  in 
which  the  triangles  meet,  is  called  the 
vertex',  the  triangular  faces  are  called 
the  lateral  faces,  and  taken  together 
they  form  the  lateral,  or  convex,  sur- 
face; the  intersections  SA,  SB,  etc.,  of 
the  lateral  faces  are  called  the  lateral 
edges;  and  the  area  of  the  lateral  sur- 
face is  called  the  lateral  area. 

615.  The  altitude  of  a  pyramid  is  the  perpendicular 
distance  from  the  vertex  to  the  plane  of  the  base. 

616.  A  pyramid  is  called  triangular,  quadrangular, 
jjentagonal,  etc.,  according  as  its  base  is  a  triangle,  quadri- 
lateral, pentagon,  etc. 

617.  A  triangular  pyramid  has  but  four  faces,  and  is 
called  a  tetraedro7i;  any  one  of  its  faces  can  be  taken  for  its 
base. 


Note.— The  six  edges  of  a  triangular  pyramid  may  be  divided  into  three 
pairs,  such  that  the  two  edges  of  a  pair  do  not  meet  each  other.  Since  each 
edge  meets  two  other  edges  at  one  vertex,  and  yet  another  two  ed^es  at  the  ad- 
joining vertex,  there  is  but  one  edge  left  to  pair  with  it.  The  pair  is  called  a 
pair  of  opposite  edges. 

618.  A  regular  jjyranud  is  one  whose  base  is  a  regular 
polygon,  the  centre  of  which  coincides  with  the  foot  of  the 


BOOK   VIL— PYRAMIDS.  303 

perpendicular  let  fall  on  it  from  the  vertex.     This  perpen- 
dicular is  called  the  axis  of  the  pyramid. 

619.  The  lateral  edges  of  a  regular  pyra- 
mid are  equal  (49G);  therefore  the  lateral 
faces  are  equal  isosceles  triangles. 

620.  The  slant  height  of  a  regular  pyra- 
mid is  the  perpendicular  distance  from  the 
vertex  to  the  base  of  any  one  of  its  lateral 
faces. 

621.  Afrustic77i  of  a  pyramid  is  the  portion  of  a  pyra- 
mid included  between  its  base  and  a  plane  parallel  to  the 
base,  and  cutting  all  the  lateral  edges.  If  the  cutting 
plane  is  not  parallel  to  the  base,  the  portion  included  be- 
tween the  base  and  plane  is  called  a  truncated  pyramid. 

622.  The  altitude  of  a  frustum  is  the  perpendicular 
distance  between  its  bases. 

In  a  frustum  of  a  regular  pyramid  the  lateral  faces  are 
equal  trapezoids. 

The  slant  height  of  the  frustum  of  a  regular  pyramid  is 
the  perpendicular  distance  between  the  parallel  sides  of  one 
of  these  trapezoids. 


EXERCISES. 

1.  Show  that  the  slant  height  of  a  regular  pyramid  is  the 
straight  line  drawn  from  the  vertex  of  the  pyramid  to  the 
middle  point  of  any  side  of  the  base. 

2.  Show  that  the  lateral  faces  of  a  frustum  of  a  regular 
pyramid  are  equal  trapezoids. 

3.  Find  the  volume  of  a  right  triangular  prism  if  the 
height  is  24  inches  and  the  sides  of  the  base  are  24,  20,  and 
20  inches. 

4.  Find  the  volume  of  a  triangular  prism  if  the  height  is 
18  inches  and  the  sides  of  the  base  are  6,  8,  and  10 
inches. 


304  SOLID  GEOMETRY. 


Proposition   1 4.    Theorem. 

623.  If  a  pyramid  is  cut  by  a  plane  parallel  to  its  base : 

(1)  The   edges  and  the  altitude  are  S 
divided  proportionally. 

(2)  TJie  secti07i  is  a  polygon  similar 
to  the  base.  %'  f  '  X'A'X 

Hyp.  Let  S-ABCDE  be  cut  by  the 
plane  abcde  \\  to  the  base,  intersecting  ^^ 
the  lateral  edges  in  a,  b,  c,  d,  e,  and  the 
altitude  in  o.  ^ 

,^,    „  Sa       Sb  So 

(1)  Toprove        _=:g^...=_. 

Proof.  Suppose  a  plane  passed  through  S  ||  to  the  base. 
Then,  because  the  edges  and  altitude  are  cut  by  three 
II  planes, 

S^_S^_        _So 
•'•SA~SB  SO*  ^  ^^^ 

(2)  To  prove  section  abode  similar  to  ABCDE. 

Proof.  Because  ab  is  ||  to  AB,  bc\\io  BC,  6T/||to  CD, 
etc.,  (519) 

.-.  labc  =  Z ABC,  Ibcd  =  ZBCD,  etc.  (525) 

Also,  since  the  A  s  Sab,  Sbc,  etc.,  are  similar  respectively 
to  the  As  SAB,  SBC,  etc.,  being  mutually  equiangular, 

•'•AB'SB'    ^^^BC~SB^     '''AB-BC* 


BOOKr  VII— PYRAMIDS.  305 

T    ,.,  he        cd       , 

In  like  manner,  ^^  =  7^^ ,  etc. 

.• .  the  polygons  ahcde  and  ABODE  are  mutually  equi- 
angular and  have  their  homologous  sides  proportional. 
.• .  the  section  ahcde  is  similar  to  the  base  ABODE. 

Q.E.D. 

624.  OoR.  1.  Because  ahcde  and  ABODE   are   similar 
polygons, 


ahcde  ah         Sb        So 


ABODE -Xb'"SB^~S6^ 


(379) 


Hence:  The  areas  of  paraUel  sections  of  a  pyramid  are 
proportional  to  the  squares  of  their  distances  from  the 
vertex. 

625.  OoR.  2.  If  two  pyramids  of  equal  altitudes  are 
cut  hy  planes  parallel  to  their  hases,  and  at  equal  distances 
from  their  vertices,  the  sections  will  he  proportional  to  the 
'     hases, 

3 

-,-,  ahcde         So  /nrn\ 

^°'-'  ABODE  =  gQ-  <*'''') 


Similarly, 


S'o'' 


A'B'C      g^'^ 


where  A',  B',  C,  «',  h',  c' ,  S',  0',  o',  are  the  correspond- 
ing pts.  of  the  second  pyramid. 

But  So  =  SV,  and  SO  =  S'O'. 

.  • .  ahcde  :  ABODE  =  a'h'c'  :  A'B'O'. 

626.  OoR.  3.  In  two  pyramids  of  equal  altitudes  and 
equivalent  hases,  sections  made  hy  planes  parallel  to  their 
hases  and  at  equal  distances  from  their  vertices  are  equiv- 
alent. 


306 


SOLID  GEOMETRY, 


Proposition  1  5.    Theorem. 

627.  Tiuo  triangular  pyramids  havi^ig  equivalent  bases 
and  equal  altiticdes  are  equivalent. 


Hyp.  Let  S-ABC  and  S'-A'B'C  have  equivalent  bases 
ABC,  A'B'C  in  the  same  plane  and  a  common  altitude. 

To  prove      vol.  S-ABC  =  vol.  S'-A'B'C. 

Proof.  Divide  the  common  altitude  into  any  number  of 
equal  parts,  and  through  the  pts.  of  division  pass  planes  || 
to  the  plane  of  the  bases. 

In  the  pyramid  S-ABC,  on  the  sections  DEF,  GHI,  etc., 
as  upper  bases,  construct  prisms  with  lateral  edges  =  and 
II  to  AD. 

Similarly,  construct  prisms  on  the  sections  of  S'-A'B'C, 
as  upper  bases. 

Since  the  corresponding  sections  are  equivalent,        (626) 

.' .  eacli  prism  in  S-ABC  is  equivalent  to  its  correspond- 
ing prism  in  S'-A'B'C.  (613) 


BOOK  VIL-PYBAMIBS.  .  307 

.*.  the  sum  of  the  prisms  inscribed  in  the  pyramid 
S-ABO  is  equivalent  to  the  sum  of  the  prisms  inscribed  in 
the  pyramid  S'-A'B'C. 

Now  let  the  number  of  equal  parts  into  which  the  com- 
mon altitude  is  divided  be  increased  indefinitely;  the  sum 
of  the  prisms  DFE-M,  etc.,  will  approach  the  volume  of 
the  pyramid  S-ABO  as  its  limit,  and  the  sum  of  the  prisms 
D'F'E'-M',  etc.,  will  approach  the  volume  of  the  pyramid 
S'-A'B'C  as  its  limit. 

P  .  • .  ultimately,  the  volumes  of  the  prisms  will  differ  from 
the  volumes  of  the  pyramids  by  less  than  any  assignable 
quantity. 

.-.  vol.  S-ABC  =  vol.  S'-A'B'C  q.e.d. 


Proposition  1  6.    Theorem. 

628.  The  lateral  area  of  a  7^egular  pyramid  is  equal  to 
half  the  product  of  the  perimeter  of  its  base  hy  its  slant 
height. 

Hyp,  Let  SH  be  the  slant  height  of         ^         S 
the  regular  pyramid  S-ABCDE. 

To  prove  ■  .  / 

area  S-ABCDE =i(AB-{- BO  +  etc.)SH.  //l 

Proof   Since  the  lateral  area  =  the       /J  J    ^  v-\ 
sum  of   the   equal  isosceles    As  SAB,      V  /           \   /D 
SBC,  etc.,  (619)      "\/ Y 

.  • .  area  S-ABCDE  =  J(AB  +  BC  +  etc.)SII.    (366) 

^  Q.E.D. 

629.  Cor.  The  lateral  area  of  the  frustu7n  of  a  regular 
pyrainid,  is^equal  to  half  the  sum  of  the  perimeters  of  its 
bases  mvltiplied  by  the  slant  height  of  the  frustum. 


308  •  SOLID  QEOMETliY. 


Proposition  1  7.    Theorem. 

630.  A  triangular  pyramid  is  one-tliird  of  a  triangular 
])ris7n  of  the  same  base  and  altitude. 

D*s;— -,E  Hyjj.  Let  S-ABC  be  a  triangular 

/        pyramid,  and  ABC-DSE  a  triangular 
'  prism  having  the  same  base  and  alti- 

tude. 
To  prove 
vol.  S-ABC  =  4  vol.  ABC-DSE. 
^^  Proof.  The  prism  ABC-D  is  com- 

posed   of     the    triangular    pyramid 
S-ABC  and  the  quadrangular  pyramid  S-ACED. 

Pass  the  plane  SCO  through  SC  and  SD,  dividing  the 
quadrangular  pyramid  into  the  two  triangular  pyramids 
S-ACD  and  S-ECD. 

Since  the  As  ACD  and  ECD  are  equal,  (130) 

.  • .  the  pyramids  S-ACD  and  S-ECD  have  equal  bases. 
Also,  they  have  the  same  altitude.  (615) 

.  • .  vol.  S-ACD  =  vol.  S-ECD.  (627) 

The  pyramid  S-ECD  may  be  regarded  as  having  its  ver- 
tex at  C  and  DSE  for  its  base. 

Because,  then,  the  pyramids  S-ABC  and  C-DSE  have 
equal  bases  (574),  and  the  same  altitude,  (523) 

.  • .  vol.  S-ABC  =  vol.  C-DSE.  (627) 

.  • .  the  three  pyramids  into  which  the  prism  is  divided 
are  equivalent  to  each  other. 

.  • .  vol.  S-ABC  =  1  vol.  ABC-DSE.      q.e.d. 

631.  Cor.  The  volume  of  a  triangular  pyramid  is  equal 
to  07ie'third  the  product  of  its  base  and  altitude. 


BOOK  VIL— VOLUME  OF  A  PYRAMID.  309 


Proposition  1  8.    Theorem. 

632.  The  volume  of  any  'pyramid  is  equal  to  one-third 
the  product  of  its  base  and  altitude. 

Hyp.  Let  S-ABCDE  be  any  pyramid;  8 

denote  its  volume  by  V  and  its  altitude 
byH.  . 

To  prove    Y  =  J  ABODE  X  H.  /, 

Proof  Pass  the  planes  SAC,  SAD,  /  /. 
etc.,  through  the  edge  SA  and  the  a^c-/— 
diagonals  AC,  AD,  etc.,  dividing  the  \/  "^- 
pyramid  into  triangular  pyramids.  B 

The  bases  of  these  pyramids  are  the 
A  s  which  compose  the  base  of  the  given  pyramid,  and 
their  common  altitude  is  the  altitude  H  of  the  pyramid. 

Then,  vol.  S-ABO  =  lABO  X  H, 

and  vol.  S-ACD  =  -JACD  X  H,  etc.  (631) 

But  the  volume  of  the  given  pyramid  =  the  sum  of  the 
volumes  of  the  triangular  pyramids. 

.-.  V  =  iABCDE  X  H.  Q.E.D. 

633.  Cor.  Two  pyramids  are  to  each  other  as  the  pro- 
ducts  of  their  bases  and  altitudes;  two  pyramids  having 
equivalent  bases  are  to  each  other  as  their  altitudes;  two 
pyramids  having  the  same  altitude  are  to  each  other  as 
their  bases;  two  pyramids  having  equivalent  bases  and 
equal  altitudes  are  equivalent, 

634.  ScH.  The  volume  of  any  polyedron  may  be  found 
by  dividing  it  into  pyramids,  and  computing  the  volume  of 
each  pyramid. 


310 


SOLID  GEOMETRY. 


Proposition  1 9.    Theorem. 

635.  A  frustum  of  a  triangtilnr  pyrcmiid  is  equivalent 
to  the  sum  of  three  pyramids,  having  the  same  altitude  as 
the  frustum,  and  whose  bases  are  the  lower  base,  the  upper 
base,  and  a  mea^i  proportional  betiveen  the  bases,  of  the 
frustum. 

Hyp,  Let  B  denote  the  lower  base,  b 
the  upper  base,  h  the  altitude,  and  V 
the  volume  of  the  frustum  ABC-D. 

To  prove      V  =  ^|(B  +  J  +  Vm). 

Proof  Pass  the  planes  AEC,  DEC 
through  the  pts.  A,  E,  C,  and  D,  E,  C, 
dividing  the  frustum  into  three  tri- 
angular pyramids. 

Denote  the  volume  of  E-ABC,  C-DEF,  E-ACD  by  P,  Q, 
K,  respectively. 

The  pyramid  P  has  for  its  base  the  lower  base  B  of  the 
frustum,  and  for  its  altitude  the  altitude  h  of  the  frustum. 


(1)  P  =  iB  X  h. 


(632) 


The  pyramid  Q,  has  for  its  base  the  upper  base  b  of  the 
frustum,  and  for  its  altitude  the  altitude  of  the  frustum. 

,'.{^)q  =  ^bxh,  (632) 

Sincp  the  pyramids  P  and  K,  regarded  as  having  the 
common  vertex  C,  and  their  bases,  AEB  and  AED,  in  the 
same  plane  BD,  have  a  common  altitude, 

.-.  P  :  R  =  A  AEB  :  A  AED.  (633) 


BOOK  VII.  — VOLUME  OF  A  PYRAMID.  311 

And  since  the  As  AEB,  AED  have  for  their  common 
altitude  the  altitude  of  the  trapezoid  ABED,  they  are  to 
each  other  as  their  bases.  (369) 

P  _  A  AEB  _  AB 
•'•E~  aAED~DE' 

In  like  manner  since  the  pyramids  R  and  Q,  regarded  as 
having  the  common  vertex  E,  and  their  bases,  ACD  and 
CDF,  in  the  same  plane  AF,  have  a  common  altitude, 

.•.R:Q=  aACD:  aCDF.  (633) 

And  since  the  As  ACD,  CDF  have  for  their  common 
altitude  the  altitude  of  the  trapezoid  ACFD,  they  are  to 


each  other  as  their  bases. 

(369) 

R      A  ACD      AC 
•  -Q-  aCDF~DF' 

Since            A  s  ABC  and  DEF  are  similar. 

(623) 

.-.  AB:DE-AC:DF. 

(307) 

.•.P:R  =  R:Q. 

.-.  R^  =  P  X  Q  =  {^Vi"  X  B  X  J.  From  (1)  and  (2) 

.  • .  R  =r  J7i  VWxl. 

.-.  the  third  pyramid  R  is  equivalent  to  a  pyramid  hav- 
ing the  same  altitude  h  as  the  frustum,  and  for  its  base  the 
mean  proportional  between  the  two  bases  B  and  h. 

Since  the  given  frustum  is  the  sum  of  P,  Q  and  R, 

.-.  V  =  i/i  X  B  +  V^  X  ^^  +  \h  VWxl 

=  J//(B  +  J  +  Vm).  Q.E.D. 


312 


SOLID  GEOMETRY. 


Proposition  20.    Tiieorem. 

636.  A  frustum  of  any  pyramid  is  equivalent  to  the 
sum  of  three  pyramids,  having  the  same  altitude  as  the 
frustum.,  and  luhose  bases  are  the  lower  base,  the  upper 
base,  and  a  mean  proportional  between  the  bases,  of  the 
frustum. 

Hyp.  Let  B  denote  the  lower 
base,  b  the  upper  base,  h  tlie 
altitu^le,  and  V  the  volume  of 
the  frustum  ABCD-E. 


To  prove  V  =  -  (B  +  ^  +  V^b). 

Proof.  Construct  a  triangular  f^i 
pyramid    S'-A'B'C    having    the 
same  altitude  as  S-ABCD,  and  its 
base  A'B'C  equivalent  to  ABCD,  and  in  the  same  plane 
with  it. 

Then  vol.  S-ABCD  =  vol.  S'-A'B'C.  (633) 

Produce  the  plane  EGHK  of  the  upper  base  of  the  given 
frustum  to  cut  the  pyramid  S'-A'B'C  in  E'G'H'. 

Then  EGHK  is  equivalent  to  E'G'H'.  (626) 

.  • .  vol.  S-EGHK  =  vol.  S'-E'G'H'.  (633) 

Taking  the  upper  pyramids  from  the  whole  pyramids, 

vol.  ABCD-EGHK  =  vol.  A'B'C-E'G'H'. 

But,  since  A'B'C-E'  is  a  frustum  of  a  triangular  pyra- 
mid whose  lower  and  upper  bases  and  altitude  are  B,  b, 
and  h,  respectively, 

.  • .  vol.  A'B'C-E'  =  |7i(B  +  Z*  +  VW).       (635) 
.  • .  V  =:  i^(B  -\-b-]-  Vm).       Q.E.D. 


BOOK    VU.—POLYEDBONS.  313 


Proposition  2 1 .    Theorem.* 

637.  Two  tetraedrons  loliich  have  a  triedral  angle  of  the 
one  equal  to  a  triedral  angle  of  the  other,  are  to  each  other 
as  the  products  of  the  three  edges  of  these  triedral  angles, 

C 


Hyp.  Let  S-ABC,  S-DEF  be  the  two  given  tetraedrons, 

having  the  common  triedral  Z  S  ;  and  let  Vand  V  denote 

their  volumes. 

V       SA  X  SB  X  SO 
U  prove  ^  =  ^^-^^--^^, 

Proof  Draw  CO,  FP  _L  to  the  plane  SAB. 

Then  we  may  take  the  faces  SAB,  SDE  as  the  bases, 
and  CO,  FP  as  the  altitudes,  of  the  triangular  pyramids 
C-SAB,  F-SDE. 


V       SAB  X  CO      SAB      CO 
•  •  V  -  SDE  X  FP  -  SDE  ^  FP- 

(633) 

SAB       SA  X  SB 
SDE  ~  SD  X  SE  • 

(375) 

And  since  the  rt.  As  SCO,  SEP  are  similar. 

CO     sc 
•     FP-SF* 

(307) 

V       SA  X  SB  X  sc 

SD  X  SE  X  SF  •  ^'^"^' 


314  SOLID  GEOMETRY, 


Proposition  22.    Theorem.* 

638.  A  truncated  tria7igular  prism  is  equivalent  to  the 
sum  of  three  pyramids  whose  cominon  base  is  the  loiver  base 
of  the  prism,  and  ivhose  vertices  are  the  three  vertices  of  the 
upper  base. 

Hyp,  Let  ABC-DEF  be  a  trun-  P 

cated  triangular  prism.  ^-<;<?^v/ 

To  prove   ABC-DEF    equivalent  ^.--<x<^^''  / 

to  the  sum  of  the  three  pyramids        ^t^-yp\    /    I 
E-ABC,  D-ABC,  F-ABO.  /}^/^^   / 

Proof.  Pass  the  planes  AEC  and    t^-^TV'''^^ 
DEC,  dividing  the  truncated  prism      ^\l/>^^^ 
into   the  three  pyramids    E-ABC,  b 

E-ACD,  and  E-CDF. 

The  first  pyramid  E-ABC  has  the  base  ABC  and  the 
vertex  E. 

The  second  pyramid  E-ACD  and  the  pyramid  B-ACD, 
have  the  same  base  ACD,  and  the  same  altitude,  since 
their  vertices  E  and  B  are  in  the  line  EB  ||  to  the  base 
ACD. 

.  • .  E-ACD  is  equivalent  to  B-ACD.  (633) 

But  the  pyramid  B-ACD  is  the  same  as  D-ABO. 

.  • .  E-ACD  is  equivalent  to  D-ABO. 

The  third  pyramid  E-CDF  and  the  pyramid  B-ACP, 
have  equivalent  bases  CDF  and  ACF  (367),  and  the  same 
altitude,  since  their  vertices  E  and  B  are  in  the  line  EB  ||  to 
CF,  or  to  the  plane  ACFD. 

.  • .  E-CDF  is  equivalent  to  B-ACF.  (633) 


BOOK  VII.—POLYEDRONS. 


315 


|b       But  tli^  pyramid  B-AOF  is  the  same  as  F-ABC. 
.-.  E-ODF  is  equivalent  to  F-ABC. 

.  • .  ABC-DEF  is  equivalent  to  the  sum  of  the  three  pyra- 
mids E-ABC,  D-ABC,  F-ABC.  q.e.d. 

639.  Cor.  1.  The  volume  of  a  truncated  right  triangular 
prism  is  equal  to  the  product  of  its  base  hy  one-third  the 
sum  *  of  its  lateral  edges. 


For,  the  lateral  edges  AD,  BE,  CF,  being  _L  to  the  base, 
are  the  altitudes  of  the  three  pyramids  whose  sum  is  equiva- 
lent to  the  truncated  prism.  (638) 

.-.  volume  ABC-DEF 

=  lABC  X  AD  +  lABC  X  BE  +  lABC  X  CF,  (632) 
=  ABC  X  i(AD  +  BE  +  CF). 

640.  Cor.  2.  The  volume  of  any  truncated  triangular 
prism  is  equal  to  the  product  of  its  right  section  hy  one- 
third  the  sum  of  its  lateral  edges. 

For,  the  rt.  section  GHK  divides  the 
truncated   triangular   prism   ABC-DEF        D- 
into  two  truncated  rt.  prisms  whose  vol- 
umes are 

GHKx4(AG  +  BH  +  CK),  (039)  A< 
and  GHKxKC^D-f  HE  +  KF).  B 

.  • .  their  sum  is  GHK  X  J(AD  +  BE  +  CF). 


*  Called  the  arithmetic  mean  of  its  lateral  edges,  also  its  mean  height. 


316 


SOLID  GEOMETRY, 


Similar  Polyedroks. 

641.  Similar  poly edrons  are  those  whose  corresponding 
polyedral  angles  are  equal,  and  which  have  the  same  num- 
ber of  faces,  similar  each  to  each,  and  similarly  placed. 

Homologous  faces,  edges,  aiigles,  etc.,  in  similar  poly- 
edrons  are  faces,  edges,  angles,  etc.,  which  are  similarly 
placed. 

642.  Cor.  1.  The  liomologotis  edges  of  two  similar  poly- 
edrons  are  proportional  to  each  other,  (317) 

643.  Cor.  2.  The  homologous  faces  of  two  similar  poly- 
edrons  are  proportional  to  the  squares  of  any  two  homolo- 
gous edges,  (379) 

644.  Cor.  3.  The  entire  surfaces  of  two  similar  poly- 
edrons  are  proportional  to  the  squares  of  any  two  homolo- 
gous edges,  (296) 


Proposition  23.    Theorem.* 

645.  Tico  similar  ijoly edrons  may  he  decomposed  into 
the  same  number  of  tetraedrons,  similar  each  to  each,  and 
similarly  placed. 

Hyp,  Let  ABCD:E-H  and 
abcde-h  be  two  similar  poly- 
edrons,  H  and  h  being  homolo- 
gous vertices. 

To  prove  that  they  may  be 
decomposed    into    the    same         B  C 

number  of  tetraedrons,  similar  each  to  each,  and  similaily 
placed. 

Proof,  Decompose  the  polyedron  AG  into  tetraedrons, 
by  dividing  all  the  faces  not  adjacent  to  H,  into  As,  and 
drawing  st.  lines  from  H  to  their  vertices. 


'   BOOK   VII.-SIMILAR  P0LYEDR0N8.  317 

lu  the  same  manner,  decompose  the  polyedron  ag  into 
tetraedrons  having  their  common  vertex  at  li  homologous 
toH. 

The  two  polyedrons  are  then  decomposed  into  the  same 
number  of  tetraedrons,  similarly  placed. 

In  the  two  homologous  tetraedrons  H-ABC  and  h-ahc, 
since  the  faces  AD,  HA,  HC  are  similar  respectively  to  the 
faces  ad,  lia,  lie,  (641) 

.*.  the  As  ABC,  HAB,  HBC  are  similar  respectively  to 
the  As  dbc,  hob,  hlc.  (321) 

Q-  AC!       AB        ,AH      AB  ,^^^, 

Since  =  — y-,  and  —7-  =  —:-,  (307) 

ac         ab  ah         ah  ^      ' 

,  • .  AC  :  ac  —  AH  :  ah, 

X    A    •  AC       BC        ,H0       BO  ,^^^, 

And  since         =  -. — ,  and  -,—  =  -^ — ,  (307) 

ac         be  he         be  ^       ' 

.-.  AC  :  «c;=HC:7ic. 
.  • .  the  faces  HAO  and  hac  are  similar.  (313) 

Also,  the  corresponding  triedral  Z  s  of  these  tetraedrons 
are  equal.  (567) 

.•.  the  tetraedrons  H-ABC,  h-abc  are  similar.  (641) 

In  the  same  way  it  may  be  proved  that  any  two  homol- 
ogous tetraedrons  are  similar. 

.*.  the  two  similar  polyedrons  may  be  decomposed  into 
the  same  number  of  tetraedrons,  similar  each  to  each,  and 
similarly  placed.  q.e.d. 

646.  Cor.  Any  two  homologous  lines  i7i  two  similar 
polyedrons  are  proportional  to  any  tivo  hovwlogous  edges. 


318  SOLID  GEOMETRY. 


Proposition  24.    Tlieorem. 
647.  Two  similar  tetraedrons  are  to  each  other  as  the 
cubes  of  their  homologous  edges. 

Hyp,  Let  SABC,  S'A'B'C  be  | 

two  similar  tetraedrons ;  and  let  /j\  ? 

V  and  V  denote  their  volumes.  /  I  \  /l\ 


rr  V         AB 

To  prove    v>  =  ^TgP 


A^ 


Proof,  Let  the  vertices  S  and 
S'  be  homologous.  b"  ^ 

Then,  since  the  triedral  Zs  S  and  S'  are  equal,         (641) 
V^  _      SA  X  SB  X  SC      _  _SA         SB_       SC^ 
•'•  V  ~  S'A'  X  S'B'  X  S'C  ~  S'A'  ^  S'B'  ^  S'C"  ^     '^ 
^_^__SC__AB 
■tiui  S'A' ""  S'B' ~  S'C  ~  A'B' *  ^      ^ 

V  _  AB         AB       ^  _   AB' 
'  *  •  V"'  ~  A'B'  ^  A'B'  ^  A'B'  ~  A^'^  *  '^'^'^' 

648.  Cor.  1.  7^/^o  similar  polyedrons  are  to  each  other 
as  the  cubes  of  their  homologous  edges. 

For,  two  similar  polyedrons  may  be  decomposed  into  the 
same  number  of  tetraedrons,  similar  each  to  each,  and 
similarly  placed  (645);  and  any  two  homologous  tetra- 
edrons are  to  each  other  as  the  cubes  of  their  homologous 
edges  (647),  or  as  the  cubes  of  any  two  homologous  edges 
of  the  polyedrons  (646).  Therefore  the  polyedrons  them- 
selves are  to  each  other  as  the  cubes  of  their  homologous 
edges.  (296) 

649.  Cor.  2.  Similar  prisms,  or  pyramids,  are  to  each 
other  as  the  cubes  of  their  altitudes, 

650.  Cor.  3.  Similar  polyedrons  are  to  each  other  as 
the  cubes  of  any  two  ho^nologous  lines. 


BOOK   VII.— REGULAR  FOLYEDRONS.  319 


Regular  Polyedrons. 

651.  A  rcfjnlar  polyedron  is  one  whose  faces  are  all 
equal  regular  polygons,  and  whose  polyedral  angles  are  all 
equal. 

Proposition  25.    Theorem.* 

652.  There  can  he  only  five  regular  convex  poly edrons. 

Proof.  At  least  three  faces  are  necessary  to  form  a  poly- 
edral angle,  and  the  sum  of  its  face-angles  must  be  less 
than  360°.  (566) 

1.  Because  the  angle  of  an  equilateral  triangle  is  60^, 
each  convex  polyedral  angle  may  have  3,  4,  or  5  equilateral 
triangles.  It  cannot  have  6  faces,  because  the  sum  of  6 
such  angles  is  360°,  reaching  the  limit.  Therefore  no  more 
than  three  regular  convex  polyedrons  can  be  formed  with 
equilateral  triangles  ;  the  tetraedron,  octaedron,  and  icosa- 
edron. 

2.  Because  the  angle  of  a  square  is  90°,  each  convex  poly- 
edral angle  may  have  3  squares.  It  cannot  have  4  squares 
because  the  sum  of  4  such  angles  is  360°.  Therefore  only 
one  regular  convex  polyedron  can  be  formed  with  squares  ; 
the  hexaedron,  or  cube, 

3.  Because  the  angle  of  a  regular  pentagon  is  108°,  each 
convex  polyedral  angle  may  have  3  regular  pentagons.  It 
cannot  have  4  faces  because  the  sum  of  4  such  angles  is 
432°.  Therefore  only  one  regular  convex  polyedron  can 
be  formed  of  regular  pentagons;  the  dodecaedron. 

Because  the  angle  of  a  regular  hexagon  is  120°,  and  the 
angle  of  every  regular  polygon  of  more  than  6  sides  is  yet 
greater  than  120°,  therefore  there  can  be  no  regular  convex 
polyedron  formed  of  regular  hexagons  or  of  any  regular 
polygons  of  more  than  6  sides. 

Therefore  there  can  be  only  five  regular  convex  poly 
edrons. 


320 


SOLID  GEOMETRY. 


Proposition  26.    Problem.* 

653.  To  construct  the  regular  polyedrons,  having  given 
one  of  the  edges. 

Given,  the  edge  AB. 

Required,  to  construct  the  regular 
polyedrons. 

1.  The  regular  tetraedron. 

Cons,  Upon  AB  construct  the  equi- 
lateral A  ABC. 

At  its  centre  0  erect  OD  _L  to  the 
plane  ABC  so  that  AD  =  AB. 

Join  DA,  DB,  DC. 

Then  ABCD  is  a  regular  tetraedron. 

Proof,  Since  the  faces  are  each  =  the  A  ABC,  (496) 
.  • .  the  triedral  Z  s  A,  B,  C,  D  are  all  equal.  (567) 
.  • .  ABCD  is  a  regular  tetraedron. 

2.  The  regular  hexaedron,  or  cube. 
Cons,  Upon   AB  construct  the  square 

ABCD. 
Upon  the  sides  of  this  square  construct 

the  four  equal  squares  AF,  BC,  CH,  DE, 

J.  to  the  plane  ABCD. 
Then        AG  is  a  regular  hexaedron.  ^  ° 

Proof,  Since  the  faces  are  equal  squares,  (Cons.) 

.  • .  the  triedral  Z  s  A,   B,  C,   D,   E,   F,   G,   H   are   all 
equal.  (567) 

.  • ,  ABCI)-E  }s  a  regular  hexaedron,  or  cube. 


BOOK  VIl.-REGULAR  POLYEDRONS,  321 

3.  The  regular  octaedrofu 

Cons,    Upon    AB    construct    the  E 

square  ABCD.  /i^\ 

At  its  centre  0  erect  EOF  ±  to  the  /   h-K^K 

plane  ABCD,  so  that  OE  =  OP  =  OA.  a^'---^^^^^ 

Join  the  pts.  E   and  F  to  all  the  ^^\^^p^1^/^ 

vertices  of  the  square.  Wl-'y^ 

Then  E-ABCD-F  is  a  regular  octa-  ^^ 
edron. 

Proof.  The  lines   from  E  and  F  to  A,  B,  C,  D  are  all 
equal.  (496) 

And  since  the  rt.  A  s  AOE,  AOB  are  equal,  (1^4) 

.  • .  AE  =  AB. 

.  • .  the  twelve  edges  of  the  octaedron  are  all  equal,  and 
the  faces  are  eight  equal  equilateral  A  s. 

Since  the  diagonals  BD  and  EF  are  equal  and  bisect  each 
other  at  rt.  Z  s,  (Cons.) 

.  • .  BEDF  is  a  square  ==  ABCD. 

And  since  AO  is  i.  to  BD  and  EF, 

.  • .  AO  is  _L  to  the  plane  of  BEDF.  (500) 

. •.  pyramid  A-BEDF  =  pyramid  E-ABCD.     (633) 

.  • .  polyedral  Z  A  =  polyedral  Z  E. 

Similarly,  it  can  be  shown  that  any  other  two  polyedral 

Z  s  are  equal. 

,  • .  E-ABCD-F  is  a  regular  octaedron. 


32^ 


SOLID  GEOMETRY, 


4.  The  regular  dodecaedron. 

Cons.  Upon  AB  construct  the  regular  pentagon  ABODE, 
and  join  to  its  sides  the  sides  of  five  other  equal  pentagons, 
so  inclined  to  the  plane  of  ABODE  as  to  form  five  triedral 
Z  s  at  A,  B,  0,  D,  E. 


I  i 

There  is  then  formed  a  convex  surface  FGHIK,  etc., 
composed  of  six  equal  regular  pentagons. 

Oonstruct  a  second  convex  surface  fyhik,  etc.,  equal  to 
the  first. 

Then  the  two  equal  convex  surfaces  may  be  combined  so 
as  to  form  a  single  convex  surface,  which  is  the  regular 
dodecaedron. 

Proof.  Because  the  face  Z  s  of  the  triedral  Z  s  in  FK 
are  equal  respectively  to  the  face  Zs  of  the  triedral  Zs 
in  fkf  (Cons. ) 

.  • .  the  triedral  Z  s  of  FK  and  fk  are  equal,  each  to 
each.  {b^'^i) 

Now  suppose  the  convexity  of  FK  to  be  up,  aod  the  con- 
vexity oifk  to  be  down. 

Put  the  two  surfaces  together  so  that  the  pt.  0  and  the 
side  OP  shall  coincide  with  the  pt.  n  and  the  side  no,  re- 
spectively. Then  two  consecutive  face  Z  s  of  one  surface 
will  unite  with  a  single  face  Z  of  the  other.  Thus  the 
three  faces  1, 5,  7,  will  enclose  a  triedral  Z :,  since  the  diedral 
Z  contained  by  1  and  5  is  the  diedral  Z  of  the  equal 
triedral  Z  formed  at  E. 

The  vertex  N  will  coincide  with  m,  and  a  like  triedral  Z 
will  be  formed  at  that  pt.,  and  so  of  all  the  others. 

.  • .  the  triedral  Z  s  are  all  equal,  and  the  solid  is  a  regular 
dodecaedron. 


BOOK   VIL— REGULAR  POLYEDRONS.  323 

5.  TJie  regular  icosaedron. 

Cons,  Upon  AB  construct  a  regular  pentagon  ABODE. 


At  its  centre  erect  a  _L  to  its  plane,  and  in  this  _L  take 
a  pt.  S  so  that  SA  —  AB. 

Join  SA,  SB,  SO,  SD,  SE. 

Then  S- ABODE  is  a  regular  pentagonal  pyramid  (496), 
and  each  of  its  faces  is  an  equilateral  A . 

Oomplete  the  polyedral  Z  s  at  A  and  B  by  adding  to 
each  three  equilateral  A  s  each  equal  to  SAB,  and  making 
the  diedral  Z  s  around  A  and  B  equal. 

There  is  then  formed  a  convex  surface  ODEF,  etc.,  com- 
posed of  ten  equal  equilateral  A  s. 

Oonstruct  a  second  convex  surface  cdef,  etc.,  equal  to  the 
first. 

Then  the  two  equal  convex  surfaces  may  be  combined  so 
as  to  form  a  single  convex  surface,  which  is  the  regular 
icosaedron. 

Proof.  Suppose  the  convexity  of  DG  to  be  up,  and  the 
convexity  of  dg  to  be  down. 

Put  the  two  surfaces  together  so  that  the  pt.  D,  where 
two  faces  meet,  falls  upon  the  pt.  6',  where  three  faces  meet. 
Then  two  consecutive  face  Z*s  of  one  surface  will  unite 
with  three  consecutive  face  Zsof  the  other.  Thus,  the 
two  faces  1  and  9  will  unite  with  the  three  faces  10,  11, 
12,  forming  a  polyedral  Z  of  five  faces  equal  to  S,  without 
in  any  way  changing  the  form  of  either  surface,  since  the 


324 


SOLID  GEOMETRY. 


three  diedral  Zs  contained  by  1  and  9,  10  and  11,  11  and 
12  are  those  which  belong  to  such  a  polyedral  Z . 

The  vertex  C  will  fall  at  h,  and  a  like  polyedral  /  will 
be  formed  at  that  pt.,  and  so  of  all  the  others. 

.'.  the  polyedral  /s  are  all  equal,  and  the  solid  is  a 
regular  icosaedron.  •  q.e.f. 

654.  ScH.  Models  of  the  regular  polyedrons  may  be 
easily  constructed  as  follows: 

Draw  the  following  diagrams  on  cardboard,  and  cut  them 
out.  Then  cut  half  way  through  the  board  in  the  dividing 
lines,  and  bring  the  edges  together  so  as  to  form  the  re- 
spective polyedrons. 


TETRAEDRON 


OCTAEDRON 


DODECAEDRON 


ICOSAEDROhT 


BOOK   VIL—PBOPERTIES  OF  P0LTEDR0N8.      325 
General  Properties  of  Polyedrons. 
euler's  theorem. f 
Proposition.  27.    Theorem.* 

655.  hi  any  polyedro7i  the  numher  of  edges  increased 
by  2  is  equal  to  the  whole  number  of  vertices  andfaces,l 

Hyp,  Let  E,  F,  and  V  denote  the 
number  of  edges,  faces,  and  vertices 
respectively  of  the  polyedron  S-ABCD.  > 

To  prove       E  +  2  =  V  +  F.  / 

Proof.    Beginning  with    one    face         / 
ABCD,  we  have  E  =  V.  ^"' 

If  we  annex  to  this  a  second  face 
SAB,  by  applying  one  of  its  edges,  as 
AB,  to  the  corresponding  edge  of  the  first  face,  we  form  a 
surface  having  one  edge  AB,  and  two  vertices  A  and  B 
co7}wion  to  both  faces  ;   therefore,  the  whole  number  of 
edges  is  now  one  more  than  the  whole  number  of  vertices. 
. •.  for  2  faces  ABCD  and  SAB,  E  =  V  +  1. 

If  ,we  annex  a  third  face  SBC,  adjacent  to  both  ABCD 
and  SAB,  we  form  a  new  surface  having  two  edges  SB  and 
BC,  and  three  vertices  S,  B,  C  in  common  with  the  preced- 
ing surface  ;  therefore  the  increase  in  the  number  of  edges 
is  again  one  more  than  the  increase  in  the  number  of 
vertices. 

.-.  for  3  faces,  E  =  V  +  2. 
In  like  manner,  for  4  faces,  E  =  V  -|-  3. 

And  so  on  until  every  face  but  one  has  been  annexed. 
. •.  for  (F  -  1)  faces,  E  =  V  +  F  -  2. 

But  annexing  the  last  face  adds  no  edges  nor  vertices. 
.-.  for  F  faces,  E  =  V  +  F  -  2, 
or  E  +  2  =  V  +  F.  Q.E.D. 

t  The  proof  of  this  theorem  was  first  published  by  Euler  (1752). 
%  This  proof  is  due  to  Cauchy. 


326  SOLID   GEOMETRY. 


EXERCISES, 


1.  How  many  edges  has  the  regular  tetraedron  ? 

2.  How  many  edges  has  the  regular  hexaedron  ? 

3.  The  frustum  of  a  regular  four-sided  pyramid  is  8  feet 
high,  and  the  sides  of  its  bases  are  3  feet  and  5  feet :  find 
the  height  of  an  equivalent  regular  pyramid  whose  base  is 
10  feet  square. 

Proposition  28.    Theorem.* 

656.  The  sum  of  the  face-angles  of  any  jwlyedron  is 
equal  to  four  right  angles  taken  as  ma?iy  tifues  as  the  poly- 
edral  has  vertices  less  two. 

Hyp.  Let  E,  F,  and  N  denote  the  number  of  edges,  faces, 
and  vertices  respectively,  and  S  the  sum  of  the  face  Z  s  of 
any  polyedron. 

To  prove  S  =  4  rt.  Z  s  ( V  -  2). 

Proof  Since  each  edge  is  common  to  two  faces, 

.  • .  the  whole  number  of  sides  of  the  faces  considered  as 
independent  polygons  is  2E. 

If  we  form  an  exterior  Z  at  each  vertex  of  every  polygon, 
the  sum  of  the  interior  and  exterior  Z  s  at  each  vertex  is 
2  rt.  Z  s. 

And  since  the  whole  number  of  vertices  in  the  faces  is 
2E, 

.  • .  the  sum  of  all  the  interior  and  exterior  Z  s  of  the 
faces  is  2  rt.  Z  s  X  2E,  or  4  rt.  Z  s  X  E. 

Since  the  sum  of  the  exterior  Z  s  of  each  face  is  4  rt. 
Zs,  (151) 

.' .  the  sum  of  the  ext.  Z  s  of  the  E  faces  is  4  rt.  Z  s  X  F. 
.-.  S  +  4rt.  Zs  X  E  =:4rt.  Zs  X  E. 

.•.S=r4rt.Zs(E-F). 

But  E-F  =  V-2.  (655) 

.-.  S===4rt.Zs  (V- 2).  Q.E.D, 


BOOK  VIL— EXERCISES,     THEOREMS,  i327 


exercises. 
Theokems. 

1.  Show  that  a  lateral  edge  of  a  right  prism  is  equal  to 
the  altitude. 

2.  Show  that  the  lateral  faces  of  right  prisms  are  rect- 
angles. 

3.  Show  that  every  section  of  a  prism  made  by  a  plane 
parallel  to  the  lateral  edges  is  a  parallelogram. 

4.  The  lateral  areas  of  right  prisms  of  equal  altitudes 
are  as  the  perimeters  of  their  bases. 

5.  The  opposite  faces  of  a  parallelepiped  are  equal  and 
parallel. 

6.  If  the  four  diagonals  of  a  quadrangular  prism  pass 
through  a  common  point,  the  prism  is  a  parallelepiped. 

T.  If  any  two  non-parallel  diagonal  planes  of  a  prism  are 
perpendicular  to  the  base,  the  prism  is  a  right  prism. 

See  (546). 

8.  Any  straight  line  drawn  through  the  centre  of  a 
parallelepiped,  terminating  in  a  pair  of  faces,  is  bisected  at 
that  point. 

9.  The  volume  of  a  triangular  prism  is  equal  to  the 
product  of  the  area  of  a  lateral  face  by  one-half  the  per- 
pendicular distance  of  that  face  from  the  opposite  edge. 

10.  In  a  cube  the  square  of  a  diagonal  is  three  times  the 
square  of  an  edge. 

11.  In  any  parallelepiped,  the  sum  of  the  squares  of  the 
twelve  edges  is  equal  to  the  sum  of  the  squares  of  its  four 
diagonals. 

12.  In  any  quadrangular  prism,  the  sum  of  the  squares 
of  the  twelve  edges  is  equal  to  the  sum  of  the  squares  of 
its  four  diagonals  plus  eight  times  the  square  of  the  line 
joining  the  common  middle  points  of  the  diagonals  taken 
two  and  two. 

I  yjl3.  A   plane  passing  through    a    triangular    pyramid, 


I 


328  SOLID  GEOMETRY. 

parallel  to  one  side  of  the  base  and  to  the  opposite  lateral 
edge,  intersects  its  faces  in  a  parallelogram. 

14.  The  lateral  surface  of  a  pyramid  is  greater  than  tlie 
base. 

Project  the  vertex  on  the  base,  etc. 

15.  The  four  middle  points  of  two  pairs  of  opposite 
edges  of  a  tetraedron  are  in  one  plane,  and  at  the  vertices 
of  a  parallelogram. 

(617). 

16.  The  three  lines  joining  the  middle  points  of  the 
three  pairs  of  opposite  edges  of  a  tetraedron  intersect  in  a 
point  which  bisects  them  all. 

17.  In  a  tetraedron,  the  planes  passed  through  the  three 
lateral  edges  and  the  middle  points  of  the  edges  of  the 
base  intersect  in  a  straight  line. 

The  intersections  of  the  planes  with  the  base  are  medials  of  the  base;  .* .  etc. 

18.  The  lines  joining  each  vertex  of  a  tetraedron  with 
the  point  of  intersection  of  the  medial  lines  of  the  opposite 
face  all  meet  in  a  point,  which  divides  each  line  in  the 
ratio  1  :  4. 

Note,— This  point  is  the  centre  of  gravity  of  the  tetraedron. 

19.  The  plane  which  bisects  a  diedral  angle  of  a  tetra- 
edron divides  the  opposite  edge  into  segments  which  are 
proportional  to  the  areas  of  the  adjacent  faces. 

See  (303),  (369). 

/20.  The  volume  of  a  truncated  triangular  prism  is  equal 
to  the  product  of  the  area  of  its  lower  base  by  the  perpen- 
dicular upon  the  lower  base  let  fall  from  the  intersection 
of  the  medial  lines  of  the  upper  base. 

21.  The  volume  of  a  truncated  parallelopiped  is  equal  to 
the  product  of  the  area  of  its  lower  base  by  the  perpendicu- 
lar from  the  centre  of  the  upper  base  upon  the  lower  base. 

22.  The  volume  of  a  truncated  parallelopiped  is  equal  to 
the  product  of  a  right  section  by  one-fourth  the  sum  of  its 
four  lateral  edges. 

See  (640). 

23.  Any  plane  passed  through  the  centre  of  a  parallelo- 
piped divides  it  into  two  equivalent  solids  (640). 


BOOK  VIL— NUMERICAL  EXERCISES.  329 

24.  The  portion  of  a  tetraedron  cut  off  by  a  plane  parallel 
to  one  of  its  faces  is  a  tetraedron  similar  to  the  whole 
tetraedron. 

25.  When  two  tetraedrons  have  a  diedral  angle  of  the 
one  equal  to  a  diedral  angle  of  the  other,  and  the  faces  in- 
cluding these  angles  similar  each  to  each,  and  similarly 
placed,  the  tetraedrons  are  similar. 

26.  Two  polyedrons  composed  of  the  same  number  of 
tetraedrons,  similar  each  to  each,  and  similarly  placed,  are 
similar. 

Numerical  Exekcises. 

27.  Find  the  lateral  area  of  a  right  prism  whose  altitude 
is  14  inches  and  perimeter  of  the  base  16  inches. 

\  Ans,  224  square  inches. 

28.  Find  the  volume  of* a  prism  the  area  of  whose  base  is 
24  square  inches  and  altitude  7  feet.  Ans,  VQ8  cubic  feet. 

29.  Find  the  surface  of  a  cube  whose  sides  are  each  11 
inches.  |  A7is,  726  square  inches. 

30.  Find  the  surfac£^f  a  cubical  cistern  whose  contents 
are  373,248  cubic  inches.      '  Ans\  180  square  feet. 

31.  Find  the  lateral  surface  of  a  right  prism  whose  alti- 
tude is  2  feet,  and  whose  base  is  a  regular  hexagon  of  which 
each  side  is  10  inches  long. 

32.  Find  the  depth  of  a  cubical  cistern  which  shall  hold 
1600  gallons,  each  gallon  being  231  cubic  inches. 

A71S,  5.98  feet. 

33.  If  the  dimensions  of  a  rectangular  parallelepiped  are 
20.5  feet,  18 1  feet,  and  6.75  feet,  what  is  the  edge  of  an 
equivalent  cube  ?  Ans.  11.4  feet. 

34.  Find  the  depth  of  a  cubical  box  which  shall  contain 
100  bushels  of  grain,  each  bushel  holding  2150.42  cubic 
inches.  Ans.  4.9  feet. 

35.  If  the  dimensions  of  a  rectangular  parallclopiped  are 
9,16,  and  25,  what  is  the  edge  of  an  equivalent  cube? 

36.  Find  tlie  dimensions  of  the  base  of  a  rectangular 


330  SOLID  GEOMETRY. 

parallelepiped  whose  volume  is  60,  surface  94,  and  alti- 
tude 3. 

37.  Find  the  ratio  of  two  rectangular  parallelopipeds,  if 
their  altitudes  are  each  4  feet,  and  their  bases  6  feet  by  3 
feet,  and  12  feet  by  8  feet,  respectively. 

38.  Find  the  ratio  of  two  rectangular  parallelepipeds  if 
their  dimensions  are  5,  6,  8,  and  10,  12, 16,  respectively. 

39.  Find  the  volume  of  a  right  triangular  prism,  if  the 
height  is  8  inches,  and  the  sides  of  the  base  are  6,  5,  and 

5  inches. 

40.  Find  the  lateral  area  of  a  right  pyramid  whose  slant 
height  is  4  feet,  and  whose  base  is  a  regular  octagon  of  which 
each  side  is  3  feet  long. 

41.  Find  the  volume  of  a  right  quadrangular  pyramid 
whose  altitude  is  12,  and  whose  base  is  4  feet  square. 

42.  Find  the  volume  of  a  pyramid  whose  altitude  is  20 
feet,  and  whose  base  is  a  rectangle  8  feet  by  7. 

43.  Find  the  lateral  area  of  a  right  pentagonal  pyramid 
whose  slant  height  is  18  inches,  and  each  side  of  the  base 

6  inches.  Ans.  270  square  inches. 

44.  Find  the  volume  of  a  pyramid  whose  altitude  is  20 
inches,  and  whose  base  is  a  regular  hexagon,  each  side  be- 
ing 6  inches.  Ans,  623.5386  cubic  inches. 

45.  Find  the  lateral  area  and  volume  of  a  right  triangular 
pyramid,  the  sides  of  whose  bases  are  3,  5,  and  6,  and  whose 
altitude  is  6. 

See  (398),  Ex.  2. 

46.  Find  the  volume  of  the  frustum  of  a  square  pyramid, 
the  sides  of  whose  bases  are  8  and  6  feet,  and  whose  alti- 
tude is  12  feet.  Ans.  592  square  feet. 

47.  If  a  plane  be  passed  parallel  to  the  base  of  the  pyra- 
mid in  Ex.  44,  midway  between  the  vertex  and  base,  find 
the  lateral  area  and  volume  of  the  frustum. 

48.  The  slant  height  of  the  frustum  of  a  right  pyramid 
is  6  feet,  and  the  perimeters  of  the  two  bases  are  18  feet  and 
12  feet:  what  is  the  lateral  area  of  the  frustum? 


BOOK   VlI.—NUMElilCAL  EXERCISES.  331 

49.  Find  tlio  slant  height  of  the  pyramid  whoso  frustum 
is  given  in  Ex.  48. 

50.  Find  the  volume  of  the  frustum  of  a  regular  triangu- 
lar pyramid,  the  sides  of  whose  bases  are  8  and  G,  and 
whose  lateral  edge  is  5. 

51.  A  pyramid  18  feet  high  has  a  base  containing  1G9 
square  feet.  How  far  from  the  vertex  must  a  plane  be 
passed  parallel  to  the  base  so  that  the  section  may  contain 
81  square  feet  ? 

52.  The  base  of  a  pyramid  contains  169  square  feet;  a 
plane  parallel  to  the  base  and  5  feet  from  the  vertex  cuts  a 
section  containing  81  square  feet:  find  the  height  of  the 
pyramid. 

53.  A  pyramid  16  feet  high  has  a  square  base  10  feet  on 
a  side.  Find  the  area  of  a  section  made  by  a  plane  parallel 
to  the  base  and  6  feet  from  the  vertex. 

54.  The  base  of  a  regular  pyramid  is  a  hexagon  of 
which  the  side  is  4  feet.  Find  the  height  of  the  pyramid  if 
the  lateral  area  is  eight  times  the  area  of  the  base. 

55.  F'ind  the  total  surface  of  a  regular  pyramid,  (1)  when 
each  side  of  its  square  base  is  12  feet,  and  jfche  slant  height 
is  24  feet;  (2)  when  each  side  of  its  square  base  is  30  feet, 
and  the  perpendicular  height  is  94  feet;  and  (3)  when  each 
side  of  its  triangular  base  is  8  feet,  and  the  slant  height  is 
24  feet. 

56.  Find  the  volume  of  a  regular  pyramid  when  each  side 
of  its  square  base  is  80  feet,  and  the  lateral  edge  is  202  feet. 

57.  Find  the  volume  of  a  regular  pyramid  whose  base  is 
an  equilateral  triangle  inscribed  in  a  circle  of  radius  40  feet, 
and  whose  slant  height  is  48  feet. 

58.  Find  the  lateral  edge,  lateral  area,  and  volume,  (1) 
of  a  regular  hexagonal  pyramid,  each  side  of  whose  base  is 
2,  and  whose  altitude  is  12;  (2)  of  a  frustum  of  a  regular 
triangular  pyramid,  the  sides  of  whose  bases  are  5  1^3  and 

t^3,  and  whose  altitude  is  3 ;  (3)  of  a  frustum  of  a  quadrangu- 
lar pyramid,  the  sides  of  whose  bases  are  II  and  1,  and  whose 


332  SOLID  GEOMETRY. 

altitude  is  12;  and  (4)  of  a  frustum  of  a  regular  liexagonal 
pyramid,  the  sides  of  whose  bases  are  12  and  4,  and  whose 
altitude  is  12. 

59.  Find  the  volume  of  the  frustum  of  a  regular  square 
pyramid,  the  sides  of  whose  bases  are  40  and  16  feet,  and 
whose  slant  height  is  20  feet. 

GO.  The  volume  of  a  frustum  of  a  regular  hexagonal 
pyramid  is  12  cubic  feet,  the  sides  of  the  bases  are  2  and  1 
feet :  find  the  height  of  the  frustum. 

61.  Find  the  difference  between  the  volume  of  the  frus- 
tum of  a  regular  quadrangular  pyramid,  the  sides  of  whose 
bases  are  4  and  3  feet,  and  the  volume  of  a  prism  of  the 
same  altitude,  whose  base  is  a  section  of  the  frustum  paral- 
lel to  its  bases  and  midway  between  them. 

62.  Find  the  dimensions  of  a  cube  whose  surface  shall  be 
numerically  equal  to  its  contents. 

63.  A  regular  pyramid  8  feet  high  is  transformed  into 
a  regular  prism  with  an  equivalent  base:  find  the  height  of 
the  prism. 

//64.  A  cube  whose  edge  is  3  feet  is  transformed  into  a 
right  prism  whose  base  is  a  rectangle  3  feet  by  2  feet:  find 
the  height  of  the  prism. 

65.  The  height  of  the  frustum  of  a  regular  quadrangular 
pyramid  is  12  feet,  and  the  sides  of  its  bases  are  6  and  10 
feet:  find  the  height  of  an  equivalent  regular  pyramid 
whose  base  is  24  feet  square. 

66.  A  mound  of  earth  is  raised  with  plane  sloping  sides 
and  rectangular  bases;  the  dimensions  at  the  bottom  are  80 
yards  by  10,  at  the  top  70  yards  by  1,  and  it  is  5  yards 
high:  find  its  cubical  contents. 

67.  A  bath  6  feet  deep  is  excavated;  the  area  of  the  sur- 
face at  the  top  is  100  square  yards,  at  the-  bottom  81  square 
yards:  find  the  number  of  gallons  of  water  it  will  hold. 

68.  A  railway  embankment  across  a  valley  has  the  fol- 
lowing measvires:  width  at  top  20  feet,  at  base  45  feet. 


BOOK   VIL— NUMERICAL  EXERCISES.  333 

height  11  feet,  length  at  top  1020  yards,  at  base  960  yards : 
find  its  volume. 

G9.  The  altitude  of  a  prism  is  9  feet  and  the  perimeter 
of  the  base  6  feet:  find  the  altitude  and  perimeter  of  the 
base  of  a  similar  prism  one-third  as  great. 

70.  A  pyramid  is  cut  by  a  plane  parallel  to  the  base  and 
midway  between  the  vertex  and  base:  find  the  ratio  of  the 
volume  cut  off  to  the  whole  volume. 

71.  The  length  of  one  of  the  lateral  edges  of  a  pyramid  is 
16  inches.  How  far  from  the  vertex  will  this  edge  be  cut  by 
a  plane  parallel  to  tlie  base,  which  divides  the  pyramid  into 
two  equivalant  parts  ? 

73.  The  height  of  the  frustum  of  a  pyramid  is  16  inches, 
and  two  homologous  edges  of  its  bases  are  8  and  6  inches: 
find  the  ratio  of  the  volume  of  the  frustum  to  that  of  the 
entire  pyramid. 

73.  Show  that  the  volume  of  a  regular  tetraedron  is 
equal  to  the  cube  of  its  edge  multiplied  by  -f^  V2. 

74.  Show  that  the  volume  of  a  regular  octaedron  is  equal 
to  the  cube  of  its  edge  multiplied  by  ^  V2. 

75.  Find  the  volume  of  a  regular  icosaedron  whose  edges 
are  each  20  feet. 

76.  Verify  Euler's  Theorem  in  the  case  of  all  the  regular 
polyeclrons. 

77.  Find  the  number  of  edges  in  a  pyramid,  and  in  a 
prism,  on  a  polygon  of  n  sides  as  base. 

Problems. 

78.  Given  three  indefinite  straight  lines  in  space  which 
do  not  intersect,  to  construct  a  parallelo piped  which  shall 
have  three  of  its  edges  on  these  lines. 

79.  To  cut  a  solid  tetraedral  angle  by  a  plane,  so  that 
the  section  shall  be  a  parallelogram. 


Book  VIII.* 
THE    SPHEEE.t 


Circles  of  the  Sphere  and  Tangent  Planes. 

definitions. 

657.  K  sphere  is  a  solid  bounded  by  a  curved  surface, 
every  point  of  which  is  equally  distant  from  a  point  within 
it,  called  the  centre. 

A  sphere  may  be  generated  by  the 
revolution  of  a  semicircle  ACB  about  its 
diameter  AB  as  an  axis. 

658.  A  radius  of  a  sphere  is  a 
straight  line  drawn  from  tlie  centre  to 
any  point  of  the  surface. 

A  diameter  of  a  sphere  is  a  straight 
line  passing  through  its  centre,  and  terminating  at  both 
ends  in  tlie  surface. 

All  the  radii  of  a  sphere  are  equal;  and  all  the  diameters 
are  equal,  since  each  is  the  sum  of  two  radii. 

659.  A  line  or  plane  is  tavgent  to  a  sphere  when  it  has 
but  one  point  in  common  with  the  surface  of  the  sphere. 

660.  Two  spheres  are  tangent  to  each  other  when  their 
surfaces  have  but  one  point  in  common. 

661.  The  common  point  is  called  the  j^oint  of  co7iiact, 
or  pf>int  of  tangency. 

662.  Two  spheres  are  concentric  when  they  have  the 
same  centre. 

663.  Two  spheres  are  equal  when  they  have  equal  radii. 

*  This  book  treats  of  the  properties  and  relations  of  those  parts  of  the  surface 
of  a  sphere  whieli  are  bounded  by  arcs  of  great  circles. 

t  In  teaching  tliis  subject,  as  well  as  in  teaching  Spherical  Trigonometry,  the 
class-room  should  be  furnished  with  a  spherical  black-board,  on  which  the 
student  should  draw  the  diagrams  of  spherical  surfaces. 

334 


BOOK   VIIL—jDmCLES  OF  THE  SPHERE. 


335 


Proposition  1.    Theorem. 

664.  Every  section  of  a  sphere  made  hy  a  plane  is  a  cir- 
cle. 

Hyp.  Let  AOB  be  a  plane  section 
of  the  sphere  whose  centre  is  0. 

To  prove  that  AOB  is  a  O. 

Proof.  Draw  00'  J_  to  the  plane 
AOB  meeting  it  in  0',  and  draw  OA, 
00  to  any  two  pts.  in  the  section. 

Since  OA  and  00  are  radii  of  the 
sphere, 

.•.OA  =  OC.  (658) 

And  since  OA  and  00  are  equal  oblique  lines  from  0  to 
the  plane  AOB, 

.•.0'A  =  0'C.  (498) 

But  A  and  0  are  any  two  pts.  in  the  section  AOB. 

.'.  the  curve  AOB  is  a  O  whose  centre  is  0'.     q.e.d. 

665.  Def.  The  circular  section  of  a  sphere  by  a  plane  is 
called  a  circle  of  the  sphere. 

If  the  plane  passes  through  the  centre  of  the  sphere,  the 
section  is  a  circle  whose  radius  is  equal  to  the  radius  of  the 
sphere.     This  circle  is  called  a  great  circle  of  the  sphere. 

A  section  made  by  a  plane  which  does  not  pass  through 
the  centre  of  the  sphere  is  called  a  small  circle. 

666.  Def.  The  two  points  in  which  a  diameter  of  the 
sphere,  perpendicular  to  the  plane  of  a  circle,  meets  the 
surface  of  the  sphere,  are  called  the  j^oles  of  the  circle,  and 
this  diameter  is  called  the  axis  of  the  circle;  thus,  P,  P' 
are  the  poles  of  the  circle  AOB,  and  PP'  is  its  axis.* 


P'  is  said  to  be  antipodal  to  P. 


336  SOLID  GEOMETRY. 

661.  Cor.  1.  The  line  through  the  centre  of  a  circle  of 
the  sphere,  perpendicidar  to  its  plane,  passes  through  the 
centre  of  the  sphere.  {<oQ>i) 

Therefore,  the  axis  of  a  circle  passes  through  its  centre 
{Q^^),  and  all  parallel  circles  have  the  same  axis  and  the 
same  poles. 

668.  Cor.  2.  All  great  circles  of  a  sphere  are  equal.  CoQ^) 

669.  Cor.  3.  All  sinall  circles  at  equal  distances  from 
the  centre  of  the  sphere  are  equal ;  and  of  two  circles  un- 
equally distant  from  the  centre,  the  nearer  is  the  larger, 
and  conversely. 

670.  Cor.  4.  Every  great  circle  bisects  the  sjjhere  and 
its  surface. 

For,  if  the  two  parts  are  separated,  and  then  placed  on  the 
common  base,  with  their  convexities  turned  the  same  way, 
their  surfaces  will  coincide,  since  all  the  points  of  either 
are  equally  distant  from  the  centre.  (GST) 

671.  Cor.  5.  Any  two  great  circles  lisect  each  other. 
For,  the   common    intersection   of    their   planes  passes 

through  the  centre  of  the  sphere,  and  is  a  diameter  of  each 
circle. 

672.  Cor.  G.  An  arc  of  a  great  circle  may  he  draw7i 
through  any  two  given  points  on  the  surface  of  a  sphere. 

For,  the  two  given  pts.  with  the  centre  of  the  sphere  de- 
termine the  plane  of  a  great  O  passing  through  the  two 
given  pts.  (484) 

If  the  two  pts.  are  the  extremities  of  a  diameter,  the 
position  of  the  circle  is  not  determined:  for  the  two  given 
pts.  and  the  centre  being  in  the  same  st.  line,  an  indefinite 
number  of  planes  can  be  drawn  through  them.  (482) 

673.  Cor.  7.  An  arc  of  a  circle  may  he  draimi  through 
any  three  given  points  on  the  surface  of  a  sphere. 

For,  the  three  pts.  determine  a  plane.  (484) 

Note.— By  tlie  distance  between  two  points  on  a  sphere  is  meant  the  arc  of  a 
great  circle  joining  them. 


BOOK    VIIL— CIRCLES  OF  THE  SPHERE. 


337 


Proposition  2.    Theorem. 

674.  All  points  in  the  circumference  of  a  circle  of  a 
sphere  are  equally  distant  from  each  of  its  poles. 

Hyp.  Let  P,  P'  be  the  poles  of 
the  O  ACB,  where  A,  C,  B  are  any 
pts.  on  its  Oce. 

To  prove  arcs  PA,  PC,  PB  equal. 

Proof.  Since  PP'  is  a  line  through 
the  centre  of  the  O  ACB  J_  to  its 
plane, 

.-.  the  St.  lines  PA,  PC,  PB  are  equal.         (496) 
.-.  the  arcs  PA,  PC,  PB  are  equal.  (198) 

Similarly,     the  arcs  P'A,   P'C,  P'B  are  equal.         q.e.d. 

675.  Def.  The  distance  from  any  point  in  the  circum- 
ference of  a  circle  to  its  nearest  pole  is  called  the  polar  dis- 
tance of  the  circle. 

676.  Cor.  1.  The  polar  distance  of  a  great  circle  is  a 
quadrant.  Thus,  PE,  PG  are  quadrants,  for  they  are  the 
measures  of  the  rt.  Z  s  POE,  POG,  whose  vertices  are  at 
the  centre  of  the  sphere.  (237) 

677.  Cor.  2.  If  a  point  P  on  the  surface  of  a  sjjhere  is 
at  a  quadrant^s  distance  from  the  two  points  E,  G,  of  an 
arc  of  a  great  circle,  it  is  the  pole  of  that  arc. 

For,  the  /s  POE,  POG  are  rt.  Zs;  .*.  the  radius  OP  is 
_L  to  the  plane  of  the  arc  EG  (500) ;  and  .' .  P  is  the  pole  of 
the  arc  EG.  (6G6) 

678.  ScH.  In  Spherical  Geometry,  the  term  quadrant 
usually  means  a  quadrant  of  a  great  circle. 


338  SOLID   GEOMETEY. 


Proposition  3.    Theorem. 

679.  Tlie  shortest  distajice  on  tlie  surface  of  a  sphere, 
'between  any  two  points  on  the  surface^  is  the  arc  of  a  great 
circle,  not  greater  than  a  semi-circumference,  joining  the 
two  ptoints. 

Hyp.  Let  AB  be  an  arc  of  a  great 
O,  not  greater  than  a  semiOce,  join- 
ing any  two  pts..A  and  B  on  the  sphere; 
and  let  ACEB  be  any  other  line  in  the 
surface  joining  A  and  B. 

To  prove        arc  AB  <  ACEB. 

Proof  Take  any  pt.  D  in  ACEB, 
pass  arcs  of  great  Os  through  A,  D 
and  B,  D;  and  join  0,  the  centre  of  the  sphere,  with  A,  D,  B. 

Then,  since  the  Z  s  AOB,  AOD,  DOB  are  tlie  face  Z  s  of 
the  triedral  Z  whose  vertex  is  at  0, 

.• .  Z  AOD  +  Z  DOB  >  Z  AOB.  (565) 

But  the  Z  s  AOD,  DOB,  AOB  are  measured  by  tlie  arcs 
AD,  DB,  AB,  respectively.  (236) 

.• .  arc  AD  -|-  arc  DB  >  arc  AB. 

Similarly,  joining  any  pt.  in  ACD  with  A  and  D  by  arcs 
of  great  Os,  the  sum  of  the  arcs  is  greater  than  arc  AD; 
and  joining  any  pt.  in  DEB  with  D  and  B  by  arcs  of  great 
Os,  the  sum  is  greater  than  arc  DB. 

If  this  process  is  indefinitely  repeated,  the  line  from  A 
to  B,  on  the  arcs  of  the  great  Os,  will  continually  increase 
and  approach  the  line  ACEB;  that  is,  the  sum  of  the  arcs 
of  the  great  Os  will  approach  ACEB  as  the  limit,  and  will 
always  be  greater  than  AB. 

/ .  arc  AB  <  ACEB.  q.e.d. 


BOOK  VIIL-TANQENT  PLANES. 


339 


Proposition  4.    Theorem. 

680.  A  plane  perpendicular  to  a  radius  of  a  sphere  at 
its  extremity  is  tangent  to  the  sphere. 

Hyp.  Let  the  plane  MN  be  JL  to 
the  radius  OP  at  its  extremity  P. 

To  prove  MN  tangent  to  the 
sphere. 

Proof.  Take  any  other  pt.  H  in 
the  plane,  and  join  OH. 

Becanse  the  _L  is  the  shortest  dis- 
tance from  a  point  to  a  plane,  (497) 

.-.OP  <  OH. 

.  • .  the  pt.  H  is  without  the  sphere. 
But  H  is  any  pt.  of  MN  except  P. 
.  * .  every  pt.  of  MN  except  P  is  without  the  sphere. 
.  • .  plane  MN  is  tangent  to  the  sphere  at  the  pt.  P.  (659) 

Q.E.D. 

681.  Cor.  1.  Every  straight  li?ie  perpendimtlar  to  a 
radivs  at  its  extremity  is  tangent  to  the  sphere.  (659) 

682.  Cor.  2.  Every  plane  or  line  tangent  to  a  sphere  is 
2)erpe)idicular  to  the  radius  draiun  to  the  2Joint  of  contact. 

683.  Cor.  3.  A  straight  line  tangent  to  any  circle  of  a 
sphere  lies  in  the  plane  tangent  to  the  sphere  at  the  point 
of  contact. 

684.  ScH.  1.  Any  straight  line  drawn  in  a  tangent  plane 
through  the  point  of  contact  is  tangent  to  the  sphere  at 
that  point. 

685.  ScH.  2.  Any  two  straight  lines,  tangent  to  the 
sphere  at  the  same  point,  determine  the  tangent  plane  at 
that  point. 


340  SOLID  GEOMETRY. 


Proposition  5.    Theorem.* 

686.  Tlirougli  any  four  points  not  in  the  same  plane, 
one  spherical  surface,  and  only  one,  may  pass. 

Hyp.  Let  A,  B,  0,  D  be  the  four 
pts.  not  in  the  same  plane. 

To  prove  that  one  spherical  surface, 
and  no  more,  may  pass  through  A,  B, 
C,  D. 

Proof,  Let  E,  H  be  the  centres  of 
the  Os  circumscribed  about  the  As 
BCD,  ACD,  respectively. 

Draw  EK  _L  to  plane  BCD,  and  HL  _L 
to  plane  ACD. 

Every  pt.  in  EK  is  equally  distant  from  the  pts.  B,  C,  D, 
and  every  pt.  in  HL  is  equally  distant  from  the  pts. 
A,  C,  D.  (496) 

Join  E  and  H  to  F,  the  middle  pt.  of  CD. 

EF  and  HF  are  each  J_  to  CD.  (203) 

.  • .  the  plane  through  EF  and  IIF  is  J_  to  CD,  (500) 
and  .  • .  this  plane  is  J_  to  both  planes  BCD,  ACD.  (540) 
Since  HL  is  _L  to  the  plane  ACD  at  H,  (Cons. ) 

.  • .  HL  lies  in  the  plane  EFH.  (538) 

Similarly,  it  may  be  shown  that  EK  lies  in  this  plane, 

.  • .  the  _Ls  EK,  HL  lie  in  the  same  plane ;  and,  being  J_ 
to  planes  which  are  not  || ,  cannot  be  || ,  and  .  •.  must  meet 
at  some  pt.  0. 

Since  0  is  in  the  J_s  EK  and  HL,  it  is  equally  distant 
from  B,  C,  D,  and  from  A,  C,  D.  (49G) 

.  • .  0  is  equally  distant  from  A,  B,  C,  D  ;  and  the  sphere 
described  with  0  as  a  centre  and  OA  as  a  radius,  will  pass 
through  the  pts.  A,  B,  C,  D. 


BOOK  VIIL—SrUEEE  AND   TEDRAEDRON,        341 

Also,  since  the  centre  of  any  sphere  passing  through  the 
four  pts.  A,  B,  C,  D,  must  be  in  the  _Ls  EK,  HL,        (498) 

.  • .  the  intersection  0  is  the  centre  of  the  only  sphere 
that  can  pass  through  the  four  given  pts.  q.e.d. 

687.  Cor.  1.  A  sphere  may  he  circumscribed  about  any 
tetraedron. 

688.  CoE.  2.  The  four  perpendiculars  to  the  faces  of  a 
tetraedron  through  their  centres  meet  at  the  same  point, 

689.  Cor.  3.  Tlie  six  planes  ivhich  bisect  at  right  angles 
the  six  edges  of  a  tetraedron  all  intersect  in  the  same  point* 

Proposition  6.    Theorem.* 

690.  A  sphere  may  be  inscribed  in  a  given  tetraedron. 
Hyp.  Let  ABCD  be  the  given  tetra- 
edron. 

To  prove  that  a  sphere  may  be  in- 
scribed in  ABCD. 

Proof.  Bisect  any  three  of  the  die- 
dral  Z  s  which  have  one  face  common, 
as  BC,  CD,  BD,  by  the  planes  OBC, 
OCD,  OBD,  respectively. 

Since  0  is  in  the  bisector  of  the  diedral  Z 
equally  distant  from  the  faces  ABC  and  BCD. 

In  the  same  way  0  is  equally  distant  from  the  faces  ACD 
and  BCD,  and  from  BAD  and  BCD. 

.  • .  the  pt.  0  is  equally  distant  from  the  four  faces  of  the 
tetraedron. 

.  • .  a  sphere  described  with  0  as  a  centre,  and  with  a 
radius  equal  to  the  common  distance  of  0  from  any  face, 
will  be  tangeut  to  each  face,  and  will  be  inscribed  in  the 
tetraedron.  (659) 

Q.E.D. 

691.  Cor.  The  six  planes  lohich  bisect  the  six  diedral 
angles  of  a  tetraedron  intersect  in  a  point. 


342  SOLID  GEOMETRY. 


Spherical  Triaxgles  and  Polygok^s. 
definitions. 

692.  The  angle  between  two  intersecting  curves  is  the 
angle  included  between  their  tangents  at  the  point  of 
intersection. 

When  the  two  curves  are  arcs  of  great  circles  the  angle 
is  called  a  spherical  angle, 

693.  A  spherical  2)olyg on  is  a  portion  of  the  surface  of 
a  sphere  bounded  by  three  or  more  arcs  of  great  circles. 

The  bounding  arcs  are  the  sides  of  the  polygon;  the 
points  of  intersection  of  the  sides  are  the  vertices  of  the 
polygon ;  and  the  angles  which  the  sides  make  with  each 
other  are  the  angles  of  the  polygon. 

A  diagonal  of  a  spherical  polygon  is  an  arc  of  a  great  circle 
joining  any  two  vertices  which  are  not  consecutive. 

694.  A  spherical  triangle  is  a  spherical  polygon  of  three 
sides. 

A  spherical  triangle  is  right  or  oblique,  scalene,  isosceles, 
or  equilateral,  in  the  same  cases  as  a  plane  triangle. 

Note. — Between  any  two  points  two  arcs  of  great  circles  may  be  drawn,  the 
one  less,  and  the  other  greater,  than  a  senii-circunifereuce.  In  the  present 
treatise  the  arcs  less  than  a  semi-circumference  will  be  taken,  unless  otherwise 
stated. 

695.  A  spherical  pyramid  is  a  portion  of  the  sphere 
bounded  by  a  spherical  polygon  and  the  jilanes  of  tlie  sides 
of  the  polygon.  The  centre  of  the  sphere  is  the  vertex  of 
the  pyramid,  and  the  spherical  polygon  is  its  hase. 

696.  Two  spherical  polygons  are  equal  if  they  can  be 
applied  one  to  the  other,  so  as  to  coincide. 

697.  Since  the  sides  of  a  spherical  polygon  are  arcs,  they 
are  usually  expressed  in  degrees,  minutes,  and  seconds^ 


BOOK  VIIL— SPHERICAL  ANGLES.  343 

Note.— Because  the  apparent  position  of  the  heavenly  bodies  is  referred  to  an 
imaginary  spherical  surface  whose  centre  we  occupy,  the  geometry  of  the  sur- 
face of  the  sphere  early  attracted  attention.  It  cannot  be  studied  to  any  great 
extent  without  a  knowledge  of  Trigonometry,  but  a  few  important  proposi- 
tions may  be  given,  which  illustrate  this  branch  of  geometry. 


Proposition  7.    Theorem. 

/    698.  A  spherical  angle  is  measured  hy  the  arc  of  a  great 
I  circle  described  IV ith  its  vertex  as  a  pole  and  inclnded  be- 

tiveen  its  sides,  produced  if  necessary. 
Hyp.  Let  ABO,  AB'C  be  two  arcs 

of  great    Os  intersecting  at  A;  let        y^ 

Al\  AT'  be  the  tangents  to  these     / 

arcs  at  A;  and  let  OBB'  be  a  plane    / 

through  the  centre  0  _L  to  AC,  in-    I 

tersecting  the  sphere  in  the  great  O     \ 

BB'.  \^__ 

To  prove  that  the  spherical  Z  BAB'  C 

is  measured  by  the  arc  BB'. 

Proof  Since  TA  and  T'A  are  respectively  in  the  planes 
of  the  arcs  BA  and  B'A,  and  are  JL  to  their  intersection 
AC,  (210) 

.-.  Z  TAT'  =  the  diedral  Z  BACB'. 

Also,  Z  BOB'  =  the  diedral  Z  BACB',  (528) 

and  it  is  measured  by  the  arc  BB'.  (236) 

But  the  spherical  Z  BAB'  is  measured  by  Z  TAT'.  (692) 

.  • .  it  is  measured  by  the  arc  BB'.  q.e.d. 

699.  CoR.  1.  A  spherical  angle  is  equal  to  the  diedral 
angle  between  the  pla^ies  of  the  tivo  circles. 

700.  Coft.  2.  If  two  arcs  of  great  circles  cut  each  other, 
their  vertical  angles  are  equal. 

701.  Cor.  3.  The  angles  of  a  spherical  triangle  are 
equal  to  the  diedral  angles  betioeen  the  p^lanes  of  the  sides 
of  the  triangle. 


344  80LID  GEOMETRY. 


Relation  of  a  Spheeical  Polygon  to  a  Polyedkal 

Angle. 

702.  Because  the  planes  of  all  great  circles  pass  through 
the  centre  of  the  sphere,  therefore  the  planes  of  the  sides 
of  a  spherical  polygon  form  a  polyedral 

angle  at  the  centre  0  whose  face-angles 
AOB,  BOO,  etc,  are  measured  by  the 
sides  AB,  BO,  etc.,  of  the  polygon 
(236),  and  whose  diedral  angles  OA, 
OB,  etc.,  are  equal  to  the  angles  A,  B, 
etc.,  of  the  spherical  polygon  ABO, 
etc.  (699) 

We  may  therefore  speak  of  all  the 
parts  of  a  spherical  polygon  as  angles,  meaning  ther-eby  the 
face-angles,  and  the  diedral  angles  between  tlie  faces,  of 
the   polyedral   angle   whose   vertex  is    the  centre  of  the 
sphere,  and  base  the  spherical  polygon. 

It  follows,  therefore,  that  the  properties  of  a  spherical 
polygon  and  a^  polyedral  angle  are  mutually  convertible. 
Hence : 

703.  From  any  relation  proved  between  the  face,  and 
diedral,  angles  of  a  polyedral  angle,  we  may  infer  the  same 
relation  hetween  the  sides  and  angles  of  a  spherical  polygon. 

And  conversely:  From  any  relation  proved  hetween  the 
sides  and  angles  of  a  spherical  polygon,  we  may  infer  the 
same  relation  hetiveen  the  face,  and  diedral,  angles  of  a 
polyedral  angle.     Therefore: 

704.  Each  side  of  a  spherical  triangle  is  less  than  the 
sum  of  the  other  ttvo  sides.  (565) 

705.  The  sum  of  the  sides  of  a  spherical  polygon  is  less 
than  a  circumference.  (566) 

706.  Tiuo  mutually  equilateral  triangles  on  the  same, 
or  on  equal,  spheres,  are  mutually  equiangular,  and  are 
either  equal  or  symmetrical,  (567) 


BOOK   VIIL—SPUERICAL   TRIANGLES,  345 


Symmeteical  Spherical  Triangles. 

707.  Symmetrical    Spherical    Tria7igles   are  those  in 
which  the  sides  and  angles  of  the  one  are  equal  respectively 
to  the  sides  and  angles  of  the  other, 
but  arranged  in  the  reverse  order. 

Thus,  the  spherical  triangles  ABO 
and  A'B'C  are  symmetrical  when  the 
vertices  of  the  one  are  at  the  ends  of 
the  diameters  from  the  vertices  of  the 
other.  * 

The  corresponding  triedral  angles 
0-ABO  and  O-A'B'C  are  also  sym- 
metrical. (567) 

In  the  same  way,  we  may  form  two  symmetrical  polygons 
of  any  number  of  sides. 

Two  symmetrical  triangles  are  mutually  equilateral  and 
equiangular;  yet  in  general  they  cannot  be  made  to  coin- 
cide. 

Thus,  if  in  the  symmetrical  triangles 
ABO,  A'B'O',  AB  is  made  to  coincide 
with  A'B'  to  bring  the  vertex  0  upon 
the  corresponding  vertex  0',  the  two 
convex  surfaces  would  have  to  be 
brought  together.  The  triangles  are 
in  fact  right-handed  and  left-handed,  and,  though  corre- 
sponding and  equal  in  every  detail,  can  no  more  be  con- 
ceived as  superposed  on  one  another  so  as  to  occupy  the 
same  space,  ^han  the  form  of  the  right  hand  on  that  of  the 
left  hand. 

•  Antipodal. 


346  SOLID   GEOMETRY, 

Proposition  8.    Theorem. 

708.  Tioo  symmetrical  spherical  triangles  are  equivalent. 

Hyp,  Let  ABC,  A'B'C  be  two  sym- 
metrical spherical  As  with  their  ho- 
mologous vertices  diametrically  oppo- 
site each  other.  (707) 

To  prove  area  ABC  =  area  A'B'C. 

Case  I.  When  the  triangles  are  isos- 
celes. 

Proof.  Let  BA  =  BC,  and  B'A' 
=  B'C. 

If  Z  B  be  placed  on  the  equal  Z  B' 
(702),  the  convexities  of  the  sphere  being  on  the  same  side, 
the  side  BA  will  fall  on  B'C,  and  BC  on  B'A'. 

And  since    BA  =  B'C,  and  BC  =  B'A', 

.  • .  A  will  fall  on  C,  and  C  on  A'. 

.  • .  the  two  A  s  coincide  throughout  and  are  identically 
equal. 

Case  II.   When  the  triangles  are  not  isosceles. 

Proof.  Let  P  and  P'  be  the  poles  of  the  small  Os  pass- 
ing through  the  pts.  A,  B,  C,  and  A',  B',  C,  respectively.* 

Draw  the  great  0  arcs  PA,  PB,  PC,  and  P'A',  P'B',  P'C. 

Then  PA  =  PB  =  PC.  (674) 

Similarly,  P'A'  =  P'B'  =  P'C. 

Since  two  symmetric  As  are  mutually  equilateral,    (707) 
.-.  P'A'  =z  PA,  P'B'  =  PB,  P'C  =  PC. 

.-.the  AS   PAC  and  P'A'C,    PCB  and  P'C'B',  PBA 
and  P'B'A'  are  respectively  isosceles  symmetric  A  s. 

.• .  they  are  identically  equal.  (Case  I) 

Because  sum  of  As  PAC,  PCB,  PBA  =  area  ABC; 
and  sum  of  As  P'A'C,  P'C'B',  P'B'A'  =  area  A'B'C, 

.  • .  area  ABC  =  area  A'B'C,  q.e.d. 

*  The  circle  which  passes  through  the  three  points  A,  B,  C  can  only  be  a 
small  circle  of  tlie  sphere;  for  if  it  were  a  great  circle,  the  three  sides  AB,  BC, 
CA,  would  lie  in  one  plaue,  and  the  triangle  ABC  would  be  reduced  to  one  of  its 
sides. 


BOOK  VIIL- SPBEHICAL  TRIANGLES.  347 


Proposition  9.    Theorem.* 

709.  Two  triangles  on  the  smne,  or  on  equal  spheres, 
having  two  sides  and  the  included  angle  of  one  equal  re- 
spectively to  tioo  sides  and  the  included  angle  of  the  other, 
are  either  equal  or  equivalent. 

Hyp.  Let  ABO,  DEF  be  two 
A  s  having  the  side  AB  =  DE, 
the  side  AC  =  DF,  and  the 
ZA=  ZD. 

Case  I.  When  the  given  parts 
of  the  two  AS  are  arranged  in 
the  same  order.  B 

To  prove  A  ABC  =  A  DEF. 

Proof.  The  A  ABC  may  be  placed  on  the  A  DEF,  as 
in  the  corresponding  case  of  plane  A  s,  and  will  coincide 
with  it.  (104) 

Case  II.  When  the  given  parts  are  arranged  in  inverse 
order,  as  in  A  s  ABC,  DEF'. 

To  prove         as  ABC  and  DEF'  equivalent. 

Proof.  Let  the  A  DEF  be  symmetrical  with  the  A  DEF', 
having  its  sides  and  Z  s  equal  respectively  to  those  of  DEF'. 
Then  in  the  A  s  ABC,  DEF,  we  have 

AB  =^  DE,  AC  =  DF,  z  A  =  zD, 

and  the  parts  arranged  in  the  same  order. 

.  • .  A  ABC  =  A  DEF.  (Case  I) 

But  A  DEF  is  equivalent  to  A  DEF'.  (708) 

.  • .  A  ABC  is  equivalent  to  A  DEF',  Q.E.D. 


348  SOLID  GEOMETRY. 


Proposition  lO.    Theorem.* 

710.  Two  triangles  on  the  same,  or  on  equal  spheres, 
having  a  side  and  the  two  adjacent  angles  of  one  equal  re- 
spectively to  a  side  and  the  two  adjacent  angles  of  the  other, 
are  either  equal  or  equivalent. 

Proof,  One  of  the  A  s,  or  the  A  symmetric  with  it,  may 
be  applied  to  the  other,  as  in  the  corresponding  case 
of  plane  As.  (105) 

Q.E.D. 


Proposition  1  1 .    Theorem.* 

711.  Tivo  mittually  equilateral  triangles,  on  the  same, 
or  on  equal  spheres,  are  either  equal  or  equivalent. 

Proof,  They  are  mutually  ^equiangular,  and  equal  or 
symmetrical.  (706) 

.  • .  they  are  either  equal  or  equivalent.  (708) 

Q.E.D. 

712.  Cor.  1.  In  an  isosceles  spherical  triangle,  the  angles 
opposite  the  equal  sides  are  eqiiah  A 

For,  in  the  A  ABO,  let  AB  =  AC  ;  pass 
the  arc  AD  of  a  great  O  through  A  and  / 

the  mid.  pt.   of  BC  ;  then  the  A  s  ABD        / 
and  ACD   are   mutually   equilateral,  and       / 
.  • .  mutually  equiangular.  (706)    b^ — 

713.  CoR.  2.  Tlie  arc  of  a  great  circle  drawn  from  the 
vertex  of  an  isosceles  spherical  triangle  to  the  middle  of  the 
base  is  perptendicular  to  the  base,  and  bisects  the  vertical 
angle. 


BOOK  VIIL— POLAR  TRIANGLES.  349 

Polar  Triangles.    - 
.   714.  One  spherical  triangle  is  called  the  polar  triangla. 
of  a  second  spherical  triangle  when  the  sides  of  the  first 
triangle  have  their  poles  at  the  vertices  of  the  second. 

Thus,  if  A,  B,  C  are  the  poles  of  the 
arcs  of  the  great  circles  B'C,  C'A', 
A'B',  respectively,  then  A'B'C  is  the 
polar  triangle  of  ABC. 

The   great  circles,   of  which   B'C,    '^^^^^C^^^C 
C'A',  A'B'  are  arcs,  will  form  three  0^ 

other  spherical  triangles  on  the  hemisphere.  But  the  one 
which  is  the  polar  of  ABC  is  that  whose  vertex  A',  homolo- 
gous to  A,  lies  on  the  same  side  of  BC  as  the  vertex  A ; 
and  in  the  same  way  with  the  other  vertices. 

Proposition  12.    Theorem. 

715.  If  the  first  of  ttvo  spherical  triangles  is  the  polar 
triangle  of  the  seco7id,  then  the  second  is  the  polar  triangle 
of  the  first »  A' 

Hyp.  Let  A'B'C  be  the  polar  A  of  //^ 

ABC. 

To  prove  that  ABC  is  the  polar  A  of    b'^^>^ 
A'B'C. 

Proof.   Because  B  is  the  pole  of  A'C, 

.  • .  BA'  is  a  quadrant. 
Because  C  is  the  pole  of  A'B', 

.  • .  CA'  is  a  quadrant. 

.  • .  A'  is  the  pole  of  BC.  (677) 

In  like  manner,  B'  is  the  pole  of  AC,  and  C  the  pole 
of  AB. 

Also,  A  and  A'  are  on  the  same  side  of  B'C,  and  so  of 
the  other  vertices. 

.  • .  ABC  is  the  polar  A  of  A'B'C.         q.e.d. 


350 


SOLID  GEOMETRY. 


Proposition  13.    Theorem. 

716.  In  tico  polar  triangles,  each  angle  of  one  is  the  sv})- 
plement  of  the  side  opposite  to  it  in  the  other. 

Hyp,  Let  ABC,  A'B'C  be  a  pair 
of  polar  As  in  which  A,  B,  0,  and 
h! ,  B',  C  denote  the  Z  s,  and  a,  l,  c, 
and  a',  l',  c'  denote  the  sides. 
To  prove  A  =  180° -a', 

B  =  180°  -  h', 
C  =  180°  -  c', 
A'  =  180°  -  a, 
B'  =  180°  -  h, 
C  =  180°  -  c. 
Proof.  Produce  the  sides  AB,  AC,  until  they  meet  B'C 
at  D  and  H. 

Because  A  is  the  pole  of  B'C,  AD  and  AH  are  quad- 
rants. (676) 
.-.  ZAniarcDH.  (698) 
Because  B'  is  the  pole  of  AH,  arc  B'H  =  90° 
Because  C  is  the  pole  of  AD,  arc  CD  =  90°. 

.  • .  arc  B'H  +  arc  CD  =  180°. 
But  arc  B'H  +  arc  CD  =  arc  B'C  +  arc  DH. 

.  • .  arc  B'C  +  arc  DH  =  180°. 
But  arc  B'C  =  a',  (Hyp.) 

and  arc  DH    =  Z  A.  (Proved  above) 

.•.A+«'  =  180°, 
or  A  =  180°  -  a\ 

In  the  same  way  all  the  other  relations  may  be  proved. 

Q.E.D. 

717.  ScH.  Two  polar  triangles  are  also  called  stipple- 
mental  triangles. 


BOOK  VIIL-SPIIERICAL  TRIANGLES.  351 


Proposition  14.    Theorem. 

718.  The  sum  of  the  angles  of  a  spherical  triangle  is 
greater  than  two,  and  less  than  six,  right  angles. 

Hyp,    Let  A,   B,   C   denote  the 
three  Z  s  of  the  spherical  A  ABC. 
.     To  prove  A  +  B  +  0  >  180°  and 
<  540°. 

Proof.     Let  a\  h' .  c'  denote  the 
opposite   sides   respectively  of   the  _ 

polar  A  A'B'C     Then  "^  'G' 

A  =  180°  -  a',  B  =  180°  -V,Q>  =  180°  -  c\  (716) 
Adding,    A  +  B  +  0  =  540°  -  (a'  +  &'  +  o% 
Because  a* ,  V ,  c'  are  sides  of  a  spherical  A , 

.•.rt'  +  6'  +  ^'  <360°.  (705) 

.-.  A+  B-h  C  >  180°. 
Also,  because  each  Z  of  the  A  is  <  2  rt.  Z  s, 

.• .  A  +  B  +  C  <  6  rt.  Z  s  or  <  540°  q.e.d. 

719.  Cor.  A  spherical  triangle  may  have  tioo,  or  even 
three,  right  angles  ;  also  two,  or  even  three,  obtuse  angles. 

720.  If  a  spherical  triangle  has  two  right  angles,  it  is 
called  a  M-rectangular  triangle ;  and  if  a  spherical  triangle 
has  three  right  angles,  it  is  called  a  tri-rectangular  tri- 
angle. 

*  EXERCISE. 

In  a  bi-rectangular  triangle  the  sides  opposite  the  right 
angles  arc  quadrants. 


352 


SOLID   GEOMETRY. 


Proposition  15.    Theorem.* 

721.  Ttuo  7nutvally  equiangular  triangles,  on  the  same, 
or  on  equal  spheres,  are  mutually  equilateral,  and  are  either 
equal  or  equivalent. 

Hyp.  Let  the  spherical  A  s  P 
and  Q  be  mutually  equiangular. 

To  prove  that  A  s  P  and  Q  are 
mutually  equilateral,  and  either 
equal  or  equivalent. 

Proof,  Let  P'  be  the  polar  A  of  P,  and  Q'  the  polar  A 
of  Q. 

Since  P  and  Q  are  mutually  equiangular,  (IlyP-) 

.• .  their  polar  A  s  P'  and  Q'  are  mutually  equilat- 
eral. (716) 

And  since  P'  and  Q'  are  mutually  equilateral, 

.• .  they  are  mutually  equiangular.  (706) 

But  since  P'  and  Q'  are  mutually  equiangular, 

.* .  As  P  and  Q  are  mutually  equilateral.  (716) 

.'.  As  P  and  Q  are  either  equal  or  equivalent.     (711) 

Q.E.D. 

Note.— Mutually  equiangular  spherical  triangles  are  mutually  equilateral,  only 
when  the  triangles  are  on  the  same,  or  on  equal,  spheres.  When  the  spheres  are 
unequal,  the  homologous  sides  of  the  triangles  are  no  longer  equal,  but  are  pro- 
portional to  the  radii  of  their  spheres  (433) ;  the  triangles  are  then  similar,  as  in 
the  case  of  plane  triangles. 

722.  CoK.  1.  If  tivo  angles  of  a  spherical  triangle  are 
equal,  the  triangle  is  isosceles. 


BOOK  VTIL— SPHERICAL  TRIANGLES. 


353 


For,  since  the  polar  A  is  isosceles  (716),  the  Zs  opp.  the 
two  equal  sides  are  equal  (712);  .-.  the 
given  A  is  isosceles.  (716) 

723.  Cor.  2.  If  three  planes  are 
passed  through  the  centre  of  a  sphere, 
each  perpendicular  to  the  other  two,  they 
divide  the  surface  of  the  sphere  into 
eight  equal  tri-rect angular  triangles. 

(702)  and  (721) 


Proposition  1  6.    Theorem. 

724.  hi  a  spherical  tria?igle,  the  greater  side  is  opposite 
the  greater  angle,  and  conversely, 

(1)  Hyp,    Let    ABC    be    a  A  having 
ZACB>ZB. 

To  prove  AB  >  AC. 

Proof,  Draw  CD,  an  arc  of  a  great  O, 
making  Z  BCD  =  Z  B. 

Then  DB  =  DC.        (722) 

Add  AD  to  each. 

Then  AD  +  DB  =  AD  +  DC. 

But  AD  +  DC  >  AC. 

. •.  AD  +  DB  >  AC,  or  AB  >  AC. 

(2)  Hyp.  Suppose   AB  >  AC. 
To  prove  ZACB  >  ZB. 

Proof.  If  Z  ACB  =:  Z  B,  then  AB  =  AC,       (722) 

which  is  contrary  to  the  hypothesis. 
And  if     Z  ACB  <  Z  B,  then  AB  <  AC,  (Proved  above) 
which  is  also  contrary  to  the  hypothesis. 

.-.  ZACB  >  ZB.  Q.E.D. 


(704) 


S64:  SOLID  GEOMETRY. 

Relative  Areas  of  Spherical  Figures, 
definitions. 

725.  A  lune  is  a  portion  of  the  sur- 
face of  a  sphere  included  between  two 
semi-circumferences  of  great  circles ;  as 
ACBD. 

The  a7igl6  of  a  lune  is  the  angle  be- 
tween the  semi-circumferences  which 
form  its  sides;  as  the  angle  CAD,  or  the 
angle  COD.  (699) 

726.  On  the  same,  or  on  equal,  spheres,  lunes  of  equal 
angles  are  equal,  as  they  are  evidently  superposable. 

727.  A  spherical  tvedge,  or  ting u la,  is  the  part  of  a 
sphere  bounded  by  a  lune  and  the  planes  of  its  sides;  as 
AOBCD. 

The  diameter  AB  is  called  the  edge  of  the  ungula,  and 
the  lune  ACBD  is  called  its  base, 

728.  The  spJierical  excess  of  a  spherical  triangle  is  the 
excess  of  the  sum  of  its  three  angles  over  a  straight 
angle.  (718) 

Thus,  if  the  angles  of  a  spherical  triangle  are  denoted  by 
A,  B,  C,  and  its  spherical  excess  by  E,  we  have 

E  =  A  +  B  +  C- 1st.  Z. 

The  spherical  excess  of  a  spherical  polygon  is  the  excess 
of  the  sum  of  its  angles  over  as  many  straight  angles  as  it 
has  sides,  less  two. 

EXERCISES. 

1.  Each  side  of  a  tri-rectangular  spherical  triangle  is  a 
quadrant.  (720) 

2.  If  the  three  angles  of  a  spherical  triangle  are  100°, 
120°,  95°,  find  its  spherical  excess. 


BOOK  VIII.— AREAS  OF  SPHERIGAL  FIGURES.     355 

3.  Given  a  spherical  triangle  whose  sides  are  80°,  100°, 
115°;  find  the  angles  of  its  polar  triangle. 

4.  Given  a  spherical  triangle  whose  angles  are  65°,  87°, 
115°;  find  the  sides  of  its  polar  triangle. 

5.  Show  that  the  sum  of  the  angles  of  a  spherical  pen- 
tagon is  greater  than  six,  and  less  than  ten,  right  angles. 


Proposition  1  7.    Theorem. 

729.  If  t too  arcs  of  great  circles  intersect  on  the  surface 
of  a  hemisphere,  the  sum  of  the  ttco  opposite  triangles  thus 
formed  is  equivalent  to  a  lune  whose  angle  is  equal  to  the 
angle  letiveen  the  given  arcs. 

Hyp.  Let  the  arcs  ACA',  BOB'  >^^^^>>g 

intersect  on  the  surface  of  the  hemi-  /^        /^  /  jX 

sphere  CABA'B'.  /  ,_^-/:L/b'  \ 

To    prove      A  ABO   +    aA'B'O  ti^^^^^ 

equivalent  to  the  lune  CAC'B.  V   BT"  /     /        j 

Proof.    Because  the    As  A'B'C  n1/' /       y 

and  ABC  are  symmetrical,      (707)  C'^'^^^ 

.• .  area  A'B'C  =  area  ABC.  (708) 

Add  area  ABC  to  each. 

.• .  area  ABC  +  area  A'B'C  =  area  ABC  +  area  ABC 

=  area  of  lune  CAC'B. 

Q.E.D. 

730.  ScH.  It  is  evident  that  the  two  spherical  pyra- 
mids, which  have  the  triangles  ABC,  A'B'C  for  bases,  are 
together  equivalent  to  the  spherical  wedge  whose  base  is 
the  lune  CAC'B. 

EXERCISE. 

If  the  sides  of  a  triangle  are  75°,  110°,  and  130°,  show 
that  the  angles  of  its  polar  triangle  arc  respectively  105°, 
70°,  and  50°. 


356 


80LID  GEOMETRY. 


Proposition  1  8.    Theorem. 

731.  TJie  area  of  a  lune  is  to  the  surface  of  the  sphere 
as  the  angle  of  the  lune  is  to  four  right  angles. 

Hyp.  Let  ACBD  be  a  lune,  and 
ECDH  the  great  O  whose  poles  are 
A  and  B;  let  L  denote  the  area  of 
the  lune,  S  the  surface  of  the  sphere, 
and  A  the  Z  of  the  lune  in  degrees. 

To  prove  ^  =  ^o' 

.    Proof  Since  the  Z  of  the  lune  is 
measured  by  the  arc  CD,  (698) 


.-.  A  :  360°  =  arc  CD  :  ©ce  ECDH. 


(236) 


(1)  Wlien  the  arc  CD  and  the  Qce  ECDH  are  commen- 
surablc. 

Apply  the  common  measure  to  the  Oce  ECDH,  and  let 
it  be  contained  m  times  iii  CD  and  n  times  in  ECDH. 


Then, 


arc  CD  :  Oce  ECDH  =  m  :  n. 


Through  the  pts.  of  division  of  ECDH  and  the  axis  AB 
pass  arcs  of  great  Os;  they  will  divide  the  whole  surface 
of  the  sphere  into  n  equal  lunes  (726),  of  which  the  lune 
ACBD  will  contain  fn. 

.  • .  L  :  S  =  wi  ;  7^. 

.-.  L  :  S  ==  arc  CD  :  ©ce  ECDH 

=  A  :  360°. 


BOOK  VIII.- AREAS  OF  SPHERICAL  FIGURES.    357 

(2)  Wien  the  arc  CD  and  the  Qce  BCD II  are  income 
meyisurahle. 

The  proof  in  this  case  may  be  extended  as  in  (234),  (298), 
and  (356).  Q.e.d. 

732.  Cor.  1.  Two  Junes  on  the  same,  or  on  equal, 
spheres,  are  to  each  other  as  their  angles, 

733.  Cor.  2.  If  T  denote  the  area  of  the  tri-rectan- 
gular  A  (720),  8T  will  express  the  surface  of  the  whole 
sphere.  (723) 

Then,  if  the  rt.  Z.  he  taken  as  the  unit,  we  have 

L  :  8T  =  A  :  4.  (731) 

.  • .  L  =  2A  X  T. 

Hence,  if  the  right  angle  he  taken  for  the  unit  of  angles, 
the  area  of  a  lune  is  equal  to  tioice  its  angle  mutiylied  hy 
the  area  of  the  tri-rect angular  triangle. 

734.  Cor.  3.  The  volume  of  a  spherical  wedge  is  to  the 
volume  of  the  sphere  as  the  angle  of  the  lune  is  to  four 
right  angles. 

For,  the  lunes  being  equal,  the  spherical  wedges  are  also 
equal  (727) ;  hence  two  spherical  wedges  are  to  each  other 
as  the  angles  included  between  their  planes. 

735.  Cor.  4.  If  the  right  angle  he  taken  as  the  unit  of 
angles,  the  tri-rect  angular  triangle  as  the  unit  of  surfaces, 
and  the  tri-rectangular  p>ym't^id  as  the  unit  of  voluines, 
then  the  area  of  a  lune,  and  the  volume  of  an  singula, 
are  each  expressed  hy  tiuice  its  angle.  (733)  and  (734) 

EXERCISES. 

1.  What  part  of  the  surface  of  a  sphere  is  a  lune  whose 
angle  is  60°  ?  90°  ?  120°  ? 

2.  What  part  of  the  volume  of  a  sphere  is  an  ungula 
whose  angle  is  50°  ?  90°  ?  100°  ? 


358  80L1JJ   GEOMETRY. 


Proposition  1  9.    Tiieorem. 

736.  The  area  of  a  spherical  triangle  is  equal  to  its 
spherical  excess. 

Hyp.    Let   A,    B,   0    denote   the 
numerical  measures  of  the  /  s  of  the         .. 
spherical   A   ABC,  the  rt.  Z   being       / 
the  unit  of  /  s,  and  the  tri-rectangu-       i 
lar  A  the  unit  of  areas.  ^'V — -. 

7(9;j7-oyeareaABC=A4-B+C— 2. 

Proof.  Continue  any  one  side  as 
AB  so  as  to  complete  the  great  O 

ABA'B'.     Continue   the  other  two  sides  AC  and  BC  till 
they  meet  this  O  in  A'  and  B'. 

Then,  area  ABC+area  A'BC^lune  ABA'C=2A, 

and      area  ABC+area  AB'C=lune  AB'CB=2B.     (735) 

Also,  since  the  A  s  ABC  and  A'B'C  are  together  equiv- 
alent to  a  lune  whose  angle  is  C,  (729) 

.-.  area  ABC  +  area  A'B'C  =  2C. 

But  the  sum  of  the  areas  of  the  As  ABC,  A'BC,  AB'C, 
A'B'C  is  equal  to  the  area  of  the  hemisphere,  or  4. 

.  • .  adding  these  three  equations,  we  have 

2  area  ABC  +  4  =  2A  +  2B  +  2C. 

.• .  area  ABC  =  A  +  B  +  C-2.    q.e.d. 

737.  Cor.  1.  By  a  method  similar  to  that  in  (736),  in 
connection  with  (730)  and  (735),  it  may  be  p|-oved  that 

The  volume  of  a  triangular  spherical  pyramid  is  equal 
to  the  sp)herical  excess  of  its  base  {the  volume  of  the  tri- 
rectangular  i^yramid  being  the  unit  of  volume). 


BOOK  VIII.-AREA   OF  SPHERICAL  POLYGON.    359 

738.  Cor.  2.  If  the  three  vertices  of  a  triangle  are  on  a 
great  circle,  its  three  sides  must  coincide  with  that  circle, 
and  each  angle  must  equal  180°.  The  area  of  the  tri- 
angle is  then  a  hemisphere,  and  its  spherical  excess  4  rt.  /  s 
01  360°.  Therefore  the  area  of  the  surface  of  the  whole 
sphere  is  720°.     Hence 

The  area  of  a  spherical  triangle  is  to  that  of  the  surface 
of  the  sphere  as  its  spherical  excess,  in  degrees,  is  to  720°.    i""^ 


Proposition  20.    Theorem. 

739.  The  area  of  a  si)herical  polygon  is  equal  to  its 
spherical  excess. 

Hyp.  Let  K  denote  the  area,  n  the  .      3- — ^.^^ 

number  of  sides,  and  S  the  sum  of  the  y/^"'"""^^^ 

/  s  of  the  spherical  polygon  ABODE,  a-^^T ^^-^    ] 

the  rt.  Z    and  the  tri-rectangular   A  ^^\^         Jc 

being  the  units  of   Zs  and  areas  re-  ^^^/ 

spectively.  ^ 

To  2)rove  K  =  S-2(n  —  2). 

Proof  From  any  vertex,  as  A,  draw  diagonals,  dividing 
the  polygon  into  {n  —  2)  A  s. 

Then,  since  the  area  of  each  A  =  the  sum  of  its  Z  s 
minus  2  rt.  Z  s(736) ;  and  since  the  sum  of  the  Z  s  of  the 
{71  —  2)  As  =  the  sum  of  the  Z s  of  the  polygon,  or  S, 

.-.  K  =  S-2(7i-2).  Q.E.D. 


Note.— In  the  last  three  propositions,  only  the  ratios  of  the  areas  are  ex- 
pressed. If  the  absolute  area  is  required,  the  area  of  the  surface  of  the  sphere 
must  be  known. 


360  SOLID  GEOMETRY. 

exercises. 

Theorems. 

1.  The  intersection  of  the  surfaces  of  two  spheres  is  a 
circle  whose  plane  is  at  right  angles  to  the  line  joining  the 
centres  of  the  spheres,  and  whose  centre  is  in  that  line. 

2.  If  lines  be  drawn  from  any  point  of  the  surface  of  a 
sphere  to  the  ends  of  a  diameter,  they  will  form  with  each 
other  a  right  angle. 

3.  If  any  number  of  lines  in  space  pass  through  a  point, 
the  feet  of  the  perpendiculars  upon  these  lines  from  another 
point  lie  upon  the  surface  of  a  sphere,     i  u   t^nvvo-^^M 

4.  If  two  straight  lines  are  tangent  to  a  sphere  at  the 
same  point,  the  plane  of  these  lines  is  tangent  to  the  sphere. 

5.  On  spheres  of  different  radii,  mutually  equiangular 
triangles  are  similar. 

See  (721),  note. 

6.  The  sum  of  the  two  arcs  of  great  circles  drawn  from 
the  extremities  of  one  side  of  a  spherical  triangle  to  a  point 
within  it,  is  less  than  the  sum  of  the  other  two  sides. 

7.  If  from  any  point  on  the  surface  of  a  sphere  two  arcs 
of  great  circles  are  drawn  perpendicular  to  a  circumference, 
the  shorter  of  the  two  arcs  is  the  shortest  arc  that  can  be 
drawn  from  the  given  point  to  the  circumference. 

8.  Any  lune  is  to  a  tri- rectangular  triangle  as  its  angle  is 
to  half  a  right  angle. 

9.  Spherical  polygons  are  to  each  other  as  their  spherical 
excesses. 

10.  Two  oblique  arcs  drawn  from  the  same  point  to 
points  of  the  circumference  equally  distant  from  the  foot 
of  the  perpendicular  are  equal. 

11.  Of  two  oblique  arcs,  the  one  which  meets  the  cir- 
cumference at  the  greater  distance  from  the  foot  of  the 
perpendicular  is  the  longer. 


BOOK    VIII.—NUMEKICAL  EXEUC18ES.  361 

12.  The  arc.  of  a  great  circle,  tangent  to  a  small  circle, 
is  perpendicular  to  the  radius  of  the  sphere  at  the  point  of 
contact. 

13.  If  three  spheres  intersect  one  another,  their  planes 
of  intersection  intersect  in  a  right  line  pei-pendicular  to  the 
plane  containing  the  centres  of  the  spheres. 

14.  Prove  that  this  line  is  the  locus  of  points  from  which 
tangent  lines  to  the  throe  splieres  are  equal. 

15.  Through  a  fixed  point,  within  or  without  a  sphere, 
three  lines  mutually  at  right  angles  intersect  the  sphere ; 

I  prove  that  the  sum  of  the  squares  of  the  three  chords  is  con- 
stant, depending  only  on  the  radius  of  the  sphere  and  the 
distance  of  the  point  from  the  centre. 

IG.  Prove  also  that  the  sum  of  the  squares  of  the  six  seg- 
ments is  constant. 

17.  Prove  that  the  area  of  a  spherical  triangle,  each  of 
whose  angles  is  |  of  a  right  angle,  is  equal  to  the  surface 
of  a  great  circle. 

18.  If  through  a  point  0  any  secant  OPP'  is  drawn  to 
cut  a  sphere  in  P,  P',  prove  that  OP.  OP'  is  constant. 

19.  Find  the  radii  of  the  spheres  inscribed  and  circum- 
scribed to  a  regular  tetraedron. 

Numerical  Exercises. 

20.  If  the  sides  of  a  spherical  triangle  are  respectively 
65°,  112°,  and  85°,  how  many  degrees  are  there  in  each 
angle  of  its  polar  triangle  ? 

21.  If  the  angles  of  a  spherical  triangle  are  respectively 
90°,  115°,  and  70°,  how  many  degrees  are  there  in  each  side 
of  its  polar  triangle  ? 

22.  Given  the  spherical  triangle  whose  sides  are  respec- 
tively 80°,  90°,  and  140°,  to  find  the  angles  of  its  polar 
triangle. 

23.  What  part  of  the  surface  of  a  sphere  is  a  lune  whose 
angle  is  45°?  54°?  80°?  . 


S6:^  SOLID   OEOMETRT. 

24.  What  part  of  the  volume  of  a  sphere  is  an  uiigula 
whose  angle  is  72°?  36"? 

25.  If  the  angle  of  a  lune  is  50°,  find  its  area  on  a  sphere 
whose  surface  is  72  square  inches.    Ans.  10  square  inches. 

26.  Find  the  area  of  a  spherical  triangle  whose  angles 
are  respectively  75°,  100°,  and  115°,  on  a  sphere  whose  sur- 
face is  72  square  inches.  Ans,  11  square  inches. 

27.  Fmd  the  area  of  a  spherical  triangle  each  of  whose 
angles  is  70°,  on  a  sphere  whose  surface  is  144  square 
inches,  Ans.  6  square  inches. 

28.  Find  the  area  of  a  spherical  triangle  whose  angles  are 
G0°,  90°,  and  120°,  on  a  sphere  whose  surface  is  64  square 
inches.  Aus.  8  square  inches. 

29.  Find  the  area  of  a  splierical  pol\^gon  of  six  sides  each 
of  whose  angles  is  150°,  on  a  sphere  whose  surface  is  100 
square  inches.  Ans.  25  square  inches. 

30.  Find  the  area  of  a  bi-rectangular  triangle  whose 
vertical  angle  is  108°,  on  a  sphere  wliose  surface  is  100 
square  inches.  A7is.  15  square  inches. 

31.  Find  the  area  of  a  spherical  pentagon  whose  angles 
are  respectively  138°,  112°,  131°,  168°,  and  153°,  on  a  sphere 
whose  surface  is  40  square  feet.  Ans.  9  square  feet. 

32.  Find  the  area  of  a  spherical  triangle  whose  angles 
are  61°,  109°,  and  127°,  on  a  sphere  whose  surface  is  10 
square  inches.  Ans.  1.625  square  inches. 

33.  Find  the  area  of  a  spherical  quadrangle  whose  angles 
are  170°,  139°,  126°,  and  141°,  on  a  sphere  whose  surface  is 
400  square  inches. 

34.  Find  the  area  of  a  spherical  pentagon  whose  angles 
are  122°,  128°,  131°,  160°,  and  161°,  on  a  sphere  whose 
surface  is  150  square  feet. 

35.  Find  the  angles  of  an  equilateral  spherical  triangle 
whose  area  is  equal  to  that  of  a  great  circle.      Ans.  120°. 

36.  Find  the  angles  of  an  equilateral  spherical  triangle 
whose  area  is  equal  to  that  of  an  equilateral  spherical  hex- 
agon, each  of  whose  angles  is  150°.  Ans.  80°. 


BOOK  VIII. -NUMERICAL  EXERCISES,  363 

37.  Find  the  volume  of  a  triangular  spherical  pyramid, 
the  angles  of  whose  bases  are  80"",  90°,  130°;  the  volume 
of  the  sphere  being  30  cubic  inches. 

Ans.  5  cubic  inches. 

38.  Find  the  volume  of  a  quadrangular  spherical  pyra- 
mid, the  angles  of  whose  bases  are  170°,  139°,  126°,  141°; 
the  volume  of  the  sphere  being  10  cubic  inches. 

Ans.  3  cubic  inches. 

39.  Find  the  ratio  of  the  areas  of  two  spherical  triangles 
on  the  same  sphere,  the  angles  being  60°,  84°,  129°,  and 
83°,  107°,  114°  respectively. 

40.  Find  the  circumference  of  a  small  circle  of  a  sphere 
whose  diameter  is  10  inches,  the  plane  of  the  circle  being 

3  inches  from  the  centre  of  the  sphere. 

Ans,  25.133  inches. 

41.  Find  the  circumference  of  a  small  circle  of  a  sphere 
whose  diameter  is  20  inches,  the  plane  of  the  circle  being 

4  inches  from  the  centre  of  the  sphere. 

42.  Find  the  area  of  the  circle  of  intersection  of  two 
spheres,  their  radii  being  4  and  6  inches  and  the  distance 
between  their  centres  5  inches. 

43.  Find  the  area  of  a  small  circle  of  a  sphere  whose  di- 
ameter is  5  inches,  the  plane  of  the  circle  being  1  inch  from 
the  centre  of  the  sphere. 

44.  The  radii  of  two  concentric  spheres  are  4  and  6 
inches,  a  plane  is  drawn  tangent  to  the  interior  sphere: 
find  the  area  of  the  section  made  in  the  outer  sphere. 

45.  Given  two  mutually  equiangular  triangles  on  spheres 
whose  radii  are  10  and  16  inches:  find  the  ratio  of  two 
homologous  sides  of  these  triangles. 

See  (721),  note. 


**  Problems. 

46.  To  pass  a  plane  tangent  to  a  sphere  at  a  given  point 
on  the  surface  of  the  sphere. 


B64  SOLID  GEOMETRY, 

47.  To  pass  a  plane  tangent  to  a  sphere  through  a  given 
straight  line  without  the  sphere. 

48.  To  construct  on  the  spherical  blackhoard  a  spherical 
angle  of  45°;  60°;  90°;  100°;  200°. 

From  P,  the  pt.  where  the  veilex  is  to  be  placed,  with  a  quadrant  describe  an 
arc,  which  will  represent  one  side  of  the  angle  required.  From  P  as  a  pole, 
with  a  quadrant  describe  an  arc  from  the  side  before  drawn,  to  measure  the  re- 
quired angle.  Lay  off  on  this  last  arc  from  the  first  arc  the  measure  of  the  re- 
quired angle;  and  through  the  extremity  of  this  arc  and  P  pass  a  great  circle. 
For  describing  the  arcs,  the  student  can  use  a  tape  equal  in  length  to  half  a 
great  circle  of  the  sphere,  marked  off  into  180  equal  parts. 

49.  To  construct  on  the  spherical  blackboard  a  spherical 
triangle,  having  two  sides  100°  and  80°,  and  the  included 
angle  58°. 

50.  To  construct,  as  above,  a  spherical  triangle,  having 
aside  75°,  and  the  adjacent  angles  110°  and  87°. 

51.  To  construct,  as  above,  a  spherical  triangle,  having 
its  sides  150°,  100°,  80°;  also  having  its  sides  50°,  85°, 
160°. 

52.  To  construct,  as  above,  a  spherical  triangle,  having 
two  sides  120°  and  88°,  and  the  included  angle  59°;  then 
construct  its  polar  triangle. 

53.  Given  two  points  on  the  surface  of  a  sphere,  to  de- 
scribe the  great  circle  passing  through  them. 

54.  To  bisect  a  given  arc,  or  a  given  angle,  on  a  sphere. 

55.  To  draw  an  arc  of  a  great  circle  perpendicular  to  a 
spherical  arc,  from  a  given  point  without  it. 

56.  To  erect  a  perpendicular  to  a  given  arc  of  a  great 
circle  from  a  given  point  in  the  arc. 

57.  Given  three  points  on  a  sphere,  to  describe  a  small 
circle  to  pass  through  them. 

58.  To  cut  a  given  sphere  by  a  plane  passing  through  a 
given  straight  line  so  that  the  section  shall  have  a  given 
radius. 

59.  At  a  given  point  in  a  great  circle,  to  draw  an  arc  of 
a  great  circle  making  a  given  angle  with  the  first. 

60.  Through  a  given  point  on  a  sphere,  to  draw  a  great 
circle  tangent  to  a  given  small  circle. 


BOOK  VIIL-EXERCISES.    PROBLEMS.  365 

61.  Through  a  given  point  on  a  sphere,  to  draw  a  great 
circle  tangent  to  two  given  small  circles. 

62.  To  inscribe  a  circle  in  a  given  spherical  triangle,  and 
to  circumscribe  a  circle  about  the  triangle. 

63.  To  construct  a  right  spherical  triangle,  having  (1) 
a  side  about  the  right  angle  and  the  hypotenuse;  and  (2) 
an  angle  and  the  opposite  side. 

64.  To  construct  a  spherical  triangle,  having  given  (1) 
the  three  sides;  and  (2)  two  sides  and  the  included  angle. 

65.  To  describe  a  sphere  to  cut  orthogonally  two  given 
spheres. 


Book  IX.* 

THE  THEEE  BOUND  BODIES. 

740.  The  only  solids  bounded  by  curved  surfaces,  that 
are  treated  of  in  Elementary  Geometry,  are  the  cylinder, 
the  cone,  and  the  sphere,  which  are  called  the  three 
round  bodies. 

The  Cylii^der. 

definitions. 

741.  A  cylindrical  surface  is  a  surface  generated  by  the 
motion  of  a  straight  line  AB,  called  the  generatrix,  which 
constantly  touches  a  given  curve  ACDE, 

called  the  directrix,  and  remains  paralle. 
to  its  original  position.  The  different 
positions  of  the  generatrix  are  called  eJe- 
ments  of  the  surface. 

742.  A  cylinder  is  a  solid  bounded 
by  a  cylindrical  surface  and  two  parallel 
planes.  The  cylindrical  surface  is  called 
the  lateral  surface,  and  the  plane  sur- 
faces are  called  the  bases. 

The  altitude  of  a  cylinder  is  the  perpendicular  distance 
between  its  bases. 

743.  A  right  section  of  a  cylinder  is  the  section  by  a 
plane  perpendicular  to  its  elements. 

744.  A  circular  cylinder  is  a  cylinder  whose  base  is  a 
circle. 

The  axis  of  a  circular  cylinder  is  the  straight  line  joining 
the  centres  of  its  bases. 

745.  A  right  cylinder  is  one  whose  elements  are  perpen- 
dicular to  its  bases. 

*  This  book  treats  of  the  properties  and  relations  of  the  cylinder,  tlie  cone, 
and  the  sphere,  and  shows  how  to  find  the  convex  surface  and  volume  of  each 
of  these  bodies. 

360 


BOOK  IX.^TEE  CYLINDEB. 


367 


^' 


14:6.  A  right  circular  cylinder,  called  also 
a  cylinder  of  revolution,  is  generated  by  re- 
volving a  rectangle  about  one  of  its  sides. 

747.  Similar  cylinders  of  revolution  are 
those  generated  by  similar  rectangles  revolv- 
ing round  homologous  sides. 

748.  A.fa7igciit  plane  to  a  cylinder  is  a  plane  which  con- 
tains an  element  of  the  cylinder  without  cutting  the  sur- 
face. The  element  which  the  plane  contains  is  called  the 
element  of  contact. 

Any  straight  line  in  a  tangent  plane,  which  cuts  the  ele- 
ment of  contact,  is  a  tangent  line  to  the  cylinder. 

749.  A  prism  is  inscribed  in  a  cylinder,  when  its 
bases  are  inscribed  in  the  bases  of  the  cylinder  and  its  lat- 
eral edges  are  elements  of  the  cylinder. 

750.  A  prism  is  circumscribed  about  a  cylinder,  when 
its  bases  are  circumscribed  about  the  bases  of  the  cylinder. 

Proposition   1 .    Theorem. 

751.  Every  section  of  a  cylinder  made  by  a  plane  pasS" 

ing  through  an  element  is  a  parallelogram,. 

Hyp,  Let  the  plane  ABCD  pass 
through  the  element  AB  of  cylinder  EH. 

To  prove  the  section  ABCD  a  OJ. 

Proof  A  plane  passing  through  the 
element  AB  cuts  the  Oce  of  the  base  in 
a  second  pt.  D. 

Through  D  draw  DC  ||  to  AB. 

Then  DC  is  in  the  plane  BAD. 

.  • .  DC  is  an  element  of  the  cylinder. 

.  • .  DC,  being  common  to  the  plane  and  the  lateral  sur- 
face of  the  cylinder,  is  their  intersection. 

Also,  AD  is  II  to  BC.     (519)    .  •.  ABCD  is  a  co,      (124) 

Q.E.D. 

752.  Coil.  Every  sectio?i  of  a  right  cylinder  made  by  a 
plane  passing  through  an  element  is  a  rectangle. 


(68) 

(741) 


368 


SOLID   GEOMETRY. 


Proposition  2.    Theorem, 

753.  The  lateral  area  of  a  cylinder  is  equal  to  the  j^e- 
rivieter  of  a  right  section  of  a  cylinder  omiltiplied  hy  an 
element. 

Hyp.  Let  S  denote  the  lateral  area, 
P  the  perimeter  of  a  rt.  section,  and 
E  an  element  of  the  cylinder  AC. 

To  prove      S  =  P  X  E. 

Proof.  Inscribe  in  the  cylinder  a 
prism  ABCD-C,  of  any  number  of 
faces  ;  and  let  s  denote  its  lateral  area 
and  p  the  perimeter  of  its  rt.  section. 

Then,  since  each  lateral  edge  is  an 
element  of  the  cylinder. 


s  =  p  X^. 


(749) 
(591) 


Kow  let  the  number  of  lateral  faces  of  the  inscribed 
prism  be  indefinitely  increased. 

The  perimeter  of  the  right  section  of  the  prism  will  ap- 
proach the  perimeter  of  the  right  section  of  the  cylinder  as 
its  limit.  ^        (430) 

.  • .  the  lateral  area  of  the  prism  will  approach  the  lateral 
area  of  the  cylinder  as  its  limit. 

Because,  however  great  the  number  of  the  lateral  faces, 

s=pXl^, 

and  because  p  approaches  P  as  its  limit  and  s  approaches 
S  as  its  limit, 

...  S  =  P  XE.  Q.E.D, 

754.  Cor.  1.  The  lateral  area  of  a  cylinder  of  revolution 
is  equal  to  the  circumference  of  its  hase  multiplied  hy  its 
altitude. 


(43G) 
(439) 


BOOK  IX.— LATERAL  AREA   OF  A   CYLINDER.     369 

755.  Cor.  2.  If  H  denote  the  altitude  of  a  cylinder  of 
revolution,  R  the  radius  of  the  base,  S  the  lateral  area,  and 
T  the  total  area,  we  have 

S  =  2;rR  X  H, 

T  =  2;rRH  +  2;rR'  =  27rR(R  +  II). 

756.  Co.R.  3.  Let  S,  S'  denote 
the  lateral  areas  ;  T,  T'  the  total 
areas;  R,R'  the  radii  of  the  bases; 
and  H,  H'  the  altitudes  of  two 
similar  cylinders  of  revolution. 

Then,  since  the  generating  rect- 
angles are  similar,  (74?) 

^       H 
•'•  R' 


R  +  H 
H'~R'  +  H'' 


2;rRH 
27rR'H' 


R'  ^H' 


and -7?^  = 


R(R  +  H) 
R'(R'  +  H') 


R' 


R  +  H 
R'  +  H' 


R"' 

TT8  xyi 

JjM  "P'a-  ('^^) 


Therefore,  the  lateral  areas,  or  the  total  areas,  of  two 
similar  cylinders  of  revolution  are  to  each  other  as  the 
squares  of  their  altitudes,  or  as  the  squares  of  the  radii  of 
their  bases. 


EXERCISES. 

1.  Required  the  lateral  area,  and  also  the  total  area,  of  a 
cylinder  of  revolution  whose  altitude  is  25  inches  and  the 
diameter  of  whose  base  is  20  inches. 

2.  Required  the  convex  surface  of  a  right  circular  cylinder 
whose  altitude  is  16  inches,  and  diameter  of  the  base  8  inches. 

3.  Requii'ed  the  altitude  and  radius  of  the  base  of  a  right 
circular  cylinder  whose  lateral  area  is  J  as  great  as  a  similar 
cylinder  of  which  the  altitude  is  20  feet  and  diameter  of 
the  base  8  feet. 


370 


SOLID   GEOMETRY. 


(612) 
inscribed 


Proposition  3.    Theorem. 

757.  The  volume  of  a  cylinder  is  equal  to  the  product  of 
its  base  hy  its  altitude. 

Hyp.  Let  V,  B,  H  denote  the  volume 
of  the  cylinder,  the  area  of  its  base,  and 
its  altitude,  respectively. 

To  prove  V  =  B  X  H. 

'Proof  Inscribe  in  the  cylinder  a  prisni!5 
and  let  V  and  B'  denote  its  volume  and 
the  area  of  its  base. 

Then,  since  the  altitude  of  the  prism 
is  H, 

.-.  V'  =  B'  X  H. 

Now  let  the  number  of  lateral  faces  of  the 
prism  be  indefinitely  increased. 

The  base  of  the  prism  B'  will  approach  the  base  of  the 
cylinder  B  as  its  limit,  and  the  volume  of  the  prism  V 
will  approach  the  volume  of  the  cylinder  V  as  its  limit. 

Because,  however  great  the  number  of  the  lateral  faces, 
we  always  have 

V  =  B'  X  H. 
.  • .  V   =  B    X  H.  Q.E.D. 

758.  Cor.  1.  If  R  denotes  the  radius  of  the  base  of  a 
cylinder  of  revolution,  then  B  =■  nW,  (439) 

.•.V  =  ;rR'.H.  (757) 

759.  Cor.  2.  Let  V,  V  denote  the  volumes,  R,  R'  the 
radii  of  the  bases,  and  H,  H'  the  altitudes  of  two  similar 
cylinders  of  revolution. 

Then,  since  the  generating  rectangles  are  similar,    (747) 
R  _  JI 
•'•   R'~  H'' 
^  _  R'H  _  R^       H  _  IT  ^  R"  . 

•*•  V  ~  R'^H'  ~  R'''  ^  W  ~  W  ~  W  ^^  ^ 

Therefore,  the  volumes  of  siynilar  cylinders  of  revohition 
are  to  each  other  as  the  cubes  of  their  altitudes ,  or  as  the 
cubes  of  their  radii. 


BOOK  IX.— TUB  GONE. 


371 


The  Cone. 


DEFINITIONS. 

760.  A  conical  surface  is  a  surface  generated  by  the 

h 


motion  of  a  straight  line  SA,  called  the 
generatrix,  which  constantly  touches  a 
given  curve  ABCD,  called  the  directrix^ 
and  always  passes  through  a  fixed  point  S, 
called  the  vertex.  The  different  positions 
of  the  generatrix  are  called  elements  of  the 
surface. 

If  the  generatrix  extends  on  both  sides 
of  the  vertex,  the  whole  surface  consists  of 
two  portions,  S-ABCD  and  S-ahcd,  lying  on  opposite  sides 
of  the  vertex,  which  are  called  the  loiver  and  tipjjer  napjMs, 
respectively. 

761.  A  cone  is  a  solid  bounded  by  a  conical  surface  and 
a  plane  cutting  all  its  elements.  The  conical  surface  is 
called  the  lateral  surface. 

The  base  of  a  cone  is  its  plane  surface. 

The  altitude  of  a  cone  is  the  perpendicular  distance  from 
the  vertex  to  the  base. 

76^.  A  circular  cone  is  a  cone  whose  base  is  a  circle. 

The  axis  of  a  circular  cone  is  the  straight  line  joining 
the  vertex  and  the  centre  of  its  base. 

763.  A  right  cone  is  one  whose  axis  is  perpendicular  to 
its  base. 

The  cones  priucipally  treated  of  in  elementary  geometry  are  right  cones. 

764.  A  right  circular  cone  is  a  circular  cone  whose  axis 
is  perpendicular  to  its  base.  It  is  also  called 
a  cone  of  revolution,  because  it  may  be 
generated  by  the  revokition  of  a  right  tri- 
angle SO  A,  about  one  of  its  perpendicular 
sides  SO,  as  an  axis:  the  hypotenuse  SA 
generates  the  lateral  surface,  and  OA 
generates  the  base.     All  the  elements  SA, 


372  SOLID   GEOMETRY. 

SB,  etc.,  are  equal;  and  any  one  is  called  the  slant  height 
of  the  cone. 

765.  Similar  cones  of  revolution  are  those  generated  by 
similar  right  triangles  revolving  round  homologous  per- 
pendicular sides. 

766.  A  tangent  i^lane  to  a  cone  is  a  plane  which  contains 
an  element  of  the  cone  without  cutting  the  surface.  The 
element  which  the  plane  contains  is  called  the  element  of 
contact. 

Any  straight  line  in  a  tangent  plane,  which  cuts  the 
element  of  contact,  is  a  tangent  line  to  the  cone. 

767.  A  frustum  of  a  cone  is  the  portion  of  a  cone  in- 
cluded between  its  base  and  a  plane  parallel  to  the  base  and 
cutting  all  the  elements. 

If  the  cutting  plane  is  not  parallel  to  the  base,  the  portion 
included  between  the  base  and  plane  is  called  a  truncated 
cone. 

768.  The  altitude  of  a  frustum  of  a  cone 
is  the  perpendicular  distance  between  its 
bases. 

The  slant  height  of  the  frustum  of  a  cone 
of  revolution  is  the  portion  of  an  element 
included  between  the  parallel  bases  of  the 
frustum. 

769.  A  pyramid  is  inscribed  m  a  cone  when  its  base  is 
inscribed  in  the  base  of  the  cone,  and  its  lateral  edges  are 
elements  of  the  cone. 

Any  plane  which  cuts  the  cone,  determines  a  truncated 
pyramid  inscribed  in  the  truncated  cone. 

770.  A  pyramid  is  circumscribed  about  a  cone  when  its, 
base  is  circumscribed  about  the  base  of  the  cone,  and  its 
vertex  coincides  with  the  vertex  of  the  cone. 

Any  plane  which  cuts  the  cone,  determines  a  truncated 
pyramid  circumscribed  about  the  truncated  cone. 


BOOK  IX.— THE  CONE. 


373 


Proposition  4.    Theorem. 

771.  Every  section  of  a  circular-  cone  made  hy  a  plane 
parallel  to  the  base  is  a  circle. 

Hyp.  Let  the  section  ahc  of  the  circu- 
lar cone  S-ABC  be  ||  to  the  base. 

To  prove  that  ahc  is  a  O. 

Proof,  Let  SO  be  the  axis  of  the  cone, 
cutting  the  plane  of  the  1|  section  in  the 
pt.  0, 

Through  SO  and  any  elements  SxA., 
SB,  pass  planes  cutting  the  base  in  the 
radii  OA,  OB,  and  the  ||  section  in  the 
st.  lines  oa,  oh. 

Then,  since  the  planes  ahc  and  ABC  are"[| , 

to  OA,  and  oh  is  ||  to  OB. 

So  ,      oh       So 


oa  IS 


(Hyp.) 
(519) 


oa 


OA  ~  SO ' 


and 


OB  ~  SO  • 


But 


oa  __  oh 
OA~OB* 

OA  =  OB. 

oa  =  oh.  . 


(Eadii) 


Hence,  all  the  st.  lines  drawn  from  the  pt.  o  to  the  pe- 
rimeter of  the  section  ahc  are  equal. 

.*.  the  section  is  a  O.  q.e.d. 


772.  TJie  axis  of  a  circular  cone  passes  through  the  cen- 
tres of  all  the  sections  parallel  to  the  base, 

EXERCISES. 

1.  Prove  that  every  section  of  a  cone  made  by  a  plane 
passing  tlirough  its  vertex  is  a 'triangle. 

2.  Eequired  the  volume  of  a  circular  cylinder  whose  alti- 
tude is  25  inches  and  diameter  of  the  base  20  inches. 


374 


SOLID  GEOMETRY. 


Proposition  5.    Theorem. 

773.  The  lateral  area  of  a  cone  of  revolution  is  equal  to 
the  product  of  the  circumference  of  its  base  hy  half  its  slant 
height. 

Hyp.  Let  S,  0,  L  denote  the  lateral 
area  of  the  cone,  the  Oce  of  its  base,  and 
its  slant  height,  respectively. 

To  prove      S  =  0  X  |L. 

Proof  Inscribe  in  the  cone  a  pyramid 
S-ABOD,  having  a  regular  base  of  any 
number  of  sides  ;  and  let  S'  denote  its 
lateral  area,  0'  the  perimeter  of  its  base, 
and  L'  its  slant  height. 

Then,  since  the  edges  of  the  pyramid  are  elQ«ftents  of 
the  cone  (769),  the  pyramid  is  regular. 


.-.  S'  =  C'  X  iL'. 


(628) 


Now  let  the  number  of  lateral  faces  of  the  inscribed 
pyramid  be  indefinitely  increased. 

C  will  approach  C  as  its  limit.  -  (430) 

.  • .  L'  and  S'  will  appi-oach  L  and  S  respectively,  as  their 
limits. 

Because,  however  great  the  number  of  lateral  faces. 


S'  =  C  X  iL', 

.  S  =  C   X  iL. 


Q.E.D. 


774.  Cor.  \.  If  R  is  the  radius  of  the  base  of  a  cone  of 
revolution,  and  T  is  the  total  area,  we  have 

S  =  27rR  X  IL  =  ttRL.  (436) 

T  r:=  ;rRL  +  nW  =  nlX  {h -\-  R). 


BOOK  IX.-FUWTUM  OF  A    CONE.  375 

775.  CoK.  2.  By  the  process  that  was  employed  in  (756) 
we  may  show  that  the  lateral  areas,  or  the  total  areas,  of 
two  similar  cones  of  revolution  are  to  each  other  as  the 
squares  of  their  radii,  or  of  their  slant  heights,  or  of  their 
altitudes. 


Proposition  6.    Tiieorem. 

776.  The  lateral  area  of  a  frustum  of  a  cone  of  revolu- 
tion is  equal  to  half  the  sum  of  the  circumferences  of  its 
bases  multiplied  by  its  slant  height. 

Hyp.  Let  S,  C,  c,  L  denote  the  lateral 
area  of  the  frustum,  the  Oces  of  its  bases, 
and  its  slant  height,  respectively. 

To  prove  S  =  ^(C  +  c)L. 

Proof.   Inscribe  in  the  frustum    of  the 
cone  the  frustum  of  a  regular  pyramid ;  and 
let  S'  denote  its  lateral  area,  C  and  c'  the 
perimeters  of  its  lower  and  upper  bases,  respectively,  and 
L'  its  slant  height. 

Then,  S'  =  i(C'  +  c')^.  (629) 

Now  let  the  number  of  lateral  faces  of  the  inscribed  frus- 
tum be  indefinitely  increased. 

C  and  c'  will  approach  0  and  c  respectively,  as  their 
limits.  (430) 

.  • .  L'  and  S'  will  approach  L  and  S  respectively,  as  their 
limits, 

.-.  S=:i(C  +  c)L.  Q.E.D. 

777.  Cor.  TJie  lateral  area  of  a  frustum  of  a  cone  of 
revolution  is'^qual  to  the  circumference  of  a  section  eqiddis- 
tant  from  its  bases*  7nultiplied  by  its  slafit  height. 


*  Called  the  mid-section. 


376  SOLID  GEOMETRY. 


Proposition  7.    Theorem. 

778.  The  volume  of  any  cm e  is  eqital  to  one-third  the 
2)rodiict  of  its  hase  and  altitude. 

Hyp.  Let  V,  B,  H  denote  the  volume  of 
the  cone,  the  area  of  its  base,  and  its  alti- 
tude, respectively. 

To  prove         V  =  JB  X  H. 

Proof.  Inscribe  in  the  cone  a  pyramid, 
and  let  V  and  B'  denote  its  volume  and 
the  area  of  its  base. 

Then,  V  =  JB'  X  H.         (632) 

Now  let  the  number  of  lateral  faces  of  the  pyramid  be  in- 
definitely increased. 

B'  will  approach  B  as  its  limit.  (430) 

.  • .  V  will  approach  V  as  its  limit. 

.-.V^JBxH.  Q.E.D. 

779.  Cor.  1.  For  the  volume  of  a  cone  of  revolution^ 
whose  altitude  is  H  and  radius  of  the  hase  is  R,  we  have 

•V-4;rR='H.  (439) 

780.  Cor.  2.  The  volumes  of  similar  cones  of  revolution 
are  to  each  other  as  the  cubes  of  their  altitudes,  or  as  the 
C2ibes  of  the  radii  of  their  bases.  ('^59) 

EXERCISES. 

1.  Required  the  lateral  area  and  volume  of  a  right  circu- 
lar cone  whose  altitude  is  24  inches  and  radius  of  the  base 
10  inches.  A?is.  81G.82  sq.  ins.;  2513.3  cu.  iiisT 

2.  Required  the  entire  surface  of  a  right  circular  cone 
whose  altitude  is  IG  inches  and  radius  of  the  base  12  inches. 

Ans,  1206.3?  sq.  ins. 


BOOK  IX.— VOLUME  OF  A  FRUSTUM.  377 

Proposition  8.    Theorem. 

781.  The  volume  of  a  frustum  of  any  cone  is  equal  to 
the  simi  of  the  volumes  of  three  cones,  lohose  common  altitude 
is  the  altitude  of  the  frustum,  and  luhose  bases  are  the  lower 
base,  the  upper  base,  and  a  mean  proportional  between  the 
bases  of  the  frustum, 

Hijp.  Let  V,  B,  b,  H  denote  the  volume  i^^^^^T^ 

of  the  frustum,  its  bases,  and  its  altitude,  /li^ 

respectively.  /      // 

To  prove        V  =  ^(B  +  b  +  VB^).       /cr-T'^X-. 

Proof   Inscribe  in  the  frustum  of  the    ^^^^^^^'L^ 
cone  the  frustum  of  a  pyramid,  and  let 
V,  B',  b'  denote  its  volume  and  the  areas  of  its  bases. 

Then,  V  =  -?(B'  +  ^'  +  \^Wb').  (635) 

o 

Now  let  the  number  of  lateral  faces  of  the  inscribed 
frustum  be  indefinitely  increased. 

B'  and  V  will  approach  B  and  b  respectively,  as  their 
limits.  (430) 

.  • .  V  will  approach  V  as  its  limit. 

.•.V=-^(B  +  &+   I^B^).  Q.E.D. 

t) 

782.  Cor.  1.  If  the  frustum  is  that  of  a  right  circ7ilar 
cone,  and  the  radii  of  its  bases  are  R  and  r,  ice  have  B  = 
7tW,  b  =  7cr\ 

.-.  V  =  i;rH(R'  +  r=  +  Rr). 

783.  Cor.  2.  This  formula  may  be  put  into  the  form 

Note.— The  volume  of  a  cask  may  be  found  approximately  by  this  formula, 
in  which  H  =  total  height  of  cask,  R  =  radius  of  mid-section,  and  r  =  radius  of 
end. 
The  nearest  approximation  in  the  case  of  most  casks  Is  given  by  the  fornmla 
V  =  j7rn[2R2  +  ra  -  i(R«  -  r")].* 
*  S^  Rouch6  et  Comberousse,  p.  155. 


378  SOLID  GEOMETRY. 


The  Sphere. 

definitions. 

784.  A  zone  is  a  portion  of  the  surface  of  a  sphere  in- 
chided  between  two  parallel  planes. 

The  altitude  of  the  zone  is  the  perpendicular  distance 
between  the  parallel  planes. 

The  bases  of  the  zone  are  the  circumferences  of  the  circles 
which  bound  the  zone. 

If  one  of  the  parallel  planes  touches  the  sphere,  the  zone 
is  called  a  zone  of  one  base. 

785.  A  spherical  segment  is  a  portion  of  the  volume  of 
a  sphere  included  between  two  parallel  planes. 

The  altitude  of  the  segment  is  the  perpendicular  distance 
between  the  parallel  planes. 

The  bases  of  the  segment  are  the  sections  of  the  sphere 
made  by  the  parallel  planes. 

A  segment  of  one  base  is  a  segment  one  of  whose  bounding 
planes  touches  the  sphere. 

786.  A  spherical  sector  is  a  portion  of  the  volume  of  a 
sphere  generated  by  the  revolution  of  a  circular  sector  about 
a  diameter  of  the  circle. 

787.  Let  the  sphere  be  generated  by  the  revolution  of 
the    semicircle     ACDEFB     about    its  ^ 
diameter  AB  as  an  axis;  and  let  CG         ^y^ 
and  DH  be  drawn  perpendicular  to  the        / 
axis.     The   arc   CD   generates  a  zone        / 
whose  altitude  is  GH,  and  the  figure       I    ^^' 
CDHG  generates  a  spherical  segment      eV'~7 
whose  altitude   is   GH.     The  circum-           p^ 
ferences  generated  by  the  points  C  and  ° 

D  are  the  bases  of  the  zone,  and  the  circles  generated  by 
CG  and  DH  are  the  bases  of  the  segment. 


BOOK  IX.-THE  SPHERE.  379 

The  arc  AC  generates  a  zone  of  one  base,  and  the  figure 
ACG  a  spherical  segment  of  one  base. 

The  circular  sector  OEF  generates  a  spherical  sector 
whose  base  is  the  zone  generated  by  the  arc  EF;  the  other 
bounding  surfaces  are  the  conical  surfaces  generated  by  the 
radii  OE  and  OF. 

If  OF  coincides  with  OB,  the  spherical  sector  is  bounded 
by  a  conical  surface  and  a  zone  of  one  base. 

If  OE  is  perpendicular  to  OB,  the  spherical  sector  is 
bounded  by  a  plane  surface,  a  conical  surface,  and  a  zone. 

Proposition  9.    Tiieorem. 

788.  Tlie  area  generated  hy  a  straight  line  revolving 
about  an  axis  in  its  plane,  is  equal  to  the  product  of  the 
projection  of  the  line  on  the  axis  by  the  circtwiference  whose 
radius  is  the  perpendicular  erected  at  the  middle  2)oint  of 
the  line  and  terminated  by  the  axis. 

Hyp,  Let  AB  be  the  revolving  line,  G 

CD  its  projection  on  the  axis  GO,  and  EO 
the  X  at  the  mid.  pt.  of  AB  and  termi 
nating  in  the  axis. 

To  p>rove     area  AB  =  CD  X  27rEO. 
Proof.  Draw  EF_L,  and  AH  |i  to  GO. 
The  area  generated  by  AB  is  the  lateral 
area  of  a  frustum  of  a  cone  of  revolution,  whose  slant  height 
is  AB  and  axis  CD. 

.  •  •  area  AB  =  AB  X  27rEF.  (777) 

The  A  s  ABH  and  EOF  are  similar.  (316) 

. •.  AB  X  EF  =  AH  X  EO  =  CD  X  EO. 
.-.  area  AB  =  CD  X  27rEO. 

If  AB  meets  the  axis,  or  is  ||  to  it,  thus  generating  a 
conical,  or  a  cylindrical  surface,  the  result  is  the  same  from 
(773)  and  (753).  Q.E.i>, 


C 


380  SOLID  GEOMETRY, 


Proposition  lO.    Theorem. 

789.  The  area  of  the  surface  of  a  sphere  is  equal  to  the 
product  of  its  diameter  hy  the  circumference  of  a  great 
circle. 

Hyp.  Let  the  sphere  be  generated  by  the  A 

revolution  of  the  semicircle  ABDF  about            ^^^^ 
the  diameter  AF,  let  0  be  the  centre^  R  the         // 
radius,  and  denote  the  surface  of  the  sphere      ^/- 

by  s.  I 

To  prove  S  =  AF  x  2;rR.  \ 

Proof.  Inscribe  in  the  semicircle  a  regular        \ 
semi-polygon  ABCDEF,  of  any  number  of  t'^^ 

sides. 

Draw  m,  Qc,  etc.,  J_  to  AF  and  OH  1  to  AB. 

OH  bisects  AB.  (201) 

Then,  area  AB  =  Ahx  27rOH.  (788) 

Similarly,  area  BC  =  be  X  2;rOH,  and  so  on. 

In  equal  0«  equal  cJiorda  a/re  equally  distant  from,  the  centre  (206). 

Now  the  sum  of  the  projections  Kb ,  be,  etc.,  of  all  the 
sides  of  the  semi-polygon  make  up  the  diameter  AF. 

.  * .  the  entire  surface  generated  by  the  revolving  semi- 
polygon  =  AF  X  2;rOH. 

Now  let  the  number  of  sides  of  the  inscribed  semi-poly- 
gon be  indefinitely  increased. 

The  semi-perimeter  will  approach  the  semi-circumference 
as  its  limit,  and  OH  will  approach  the  radius  E  as  its 
limit.  (430) 

.  • .  the  surface  of  revolution  will  approach  the  surface  of 
the  sphere  as  its  limit. 

.-.  S  =  AF  X  2;rR.        ^  q.e.d, 


BOOK  IX.— THE  SPHERE.  381 

790.  CoK.  1.  Since  AF  =  2K, 

.-.  S  =  2Rx  27rR  =  4;rR'.  (789) 

Therefore,  the  area  of  the  surface  of  a  sphere  is  equal  to 
the  area  of  four  great  circles, 

791.  Cor.  2.  The  areas  of  the  surf.cts  of  two  spheres 
are  to  each  other  as  the  squares  of  their  radiif  or  as  the 
squares  of  their  diameters. 

792.  Cor.  3.  The  area  of  a  zone  is  equal  to  the  product 
of  its  altitude  by  the  circumfereuce  of  a  great  circle. 

For,  the  area  of  tlie  zone  generated  by  the  revolution  of 
the  arc  BD    . 

=  bd  X  27rn. 

793.  Cor.  4.  Zones  07i  the  same  sphere,  or  on  equal 
spheres,  are  to  each  other  as  their  altitudes. 

794.  Cor.  5.  Since  the  arc  AB  generates  a  zone  of  one 
base  whose  area  is 

AbX27rn=  nkb  X  AF  =  ;?AB',  (325) 

therefore,   the  area  of  a  zone  of  one  base  is  equal  to  the 
area  of  the  circle  whose  radius  is  the  chord  of  the  zone. 

795.  Cor.  6.  If  a  cylinder  is  circumscribed  about  a 
sphere,  the  total  area  of  the  cylinder  =  6;rR".  (755) 

Therefore,  the  area  of  the  surface  of  a  sphere  is  equal  to 
tivO'thirds  the  total  area  of  the  circumscribing  cylinder, 

EXERCISES 

1.  Find  the  area  of  the  surface  of  a  sphere  whose  radius 
is  4  inches. 

2,  Prove  that  two  zones  on  different  spheres  are  to  each 
otlier  as  the  products  of  their  altitudes  by  the  radii  of  the 
spheres. 


882  80LID   GEOMETRY. 


Proposition  1  1 .    Theorem. 

•    796.  Tlie  volume  of  a  sphere  is  equal  to  the  area  of  its 
surface  multiplied  hy  one-third  of  its  radius. 

hyp.  Let  V  denote  the  volume 
of  a  sphere,  S  tlie  area  of  its  sur- 
face, and  R  its  radius. 

To  prove    V  =  S  X  JR. 

Proof.  Conceive  the  whole  sur- 
face of  the  sphere  to  be  divided 
into  a  great  number  of  equal 
spherical  polygons.  Form  pyra- 
mids by  joining  the  vertices  of 
these  polygons  together  successively  and  to  the  centre  of 
the  sphere.  It  is  evident  that  these  pyramids  will  have  a 
common  altitude. 

The  volume  of  each  pyramid  =  base  X  \  altitude.    (631) 

.'.  the  sum  of  all  the  pyramids  =  sum  of  bases  X  \  al- 
titude. 

Now  let  the  number  of  spherical  polygons  be  indefinitely 
increased. 

The  sum  of  the  bases  of  all  the  pyramids  will  approach 
the  surface  of  the  sphere  as  its  limit;  and  the  altitude  of 
each  pyramid  will  approach  the  radius  of  the  sphere  as  its 
limit. 

.*.  the  sum  of  the  volumes  of  all  the  pyramids  will  ap- 
proach the  volume  of  the  sphere  as  its  limit. 

...V  =  SxiR.  Q.E.D. 

Note.— This  result  might  be  obtained  by  regarding  the  sphere  as  the  Hmit  of 
a  circumscribed  polyedron,  the  number  of  whose  faces  was  indefinitely  in- 
creased. For,  if  pyramids  ai-e  formed  having  the  faces  of  the  polyedron  as 
their  bases,  and  the  centre  of  the  sphere  as  their  common  vertex,  these  pyra- 
mids will  have  a  common  altitude  equal  to  the  radius  of  the  sphere. 

Then  each  pyramid  =  face  X  3  altitude. 

.-.  sum  of  pyramids  =  sum  of  faces  X  \  altitude. 

But  in  the  limit  sum  of  faces  of  polyedron  =  surface  of  sphere:  and  sum  of 
volumes  of  pyramids  =  volume  of  sphere. 

.-.  V  =  SX§R. 


BOOK  IX.— THE  SPHERE.  388 

797.  CoK.  1.  The  volume  of  a  spherical  pyrafnid  is 
equal  to  the  area  of  its  base  multij^lied  by  one-thii'd  the  ra- 
dius of  the  sphere.  (79G) 

798.  Cor.  2.  Since   S  =  47rR%  (790) 

.•.V  =  |;rR^;  (79G) 

or,  if  D  denotes  the  diameter, 

V  =  \n^\ 

799.  Cor.  3.  Tlie  volumes  of  tivo  s2Jheres  are  to  each 
other  as  the  cubes  of  their  radii. 

800.  Cor.  4.  I'iie  volume  of  a  spherical  sector  is  equal 
to  the  area  of  the  zone  tvhich  forms  its  base  multiplied  by 
one-third  the  radius  of  the  sphere. 

For,  a  spherical  sector,  like  the  entire  sphere,  may  he 
conceived  as  consisting  of  an  indefinitely  great  number  of 
pyramids  wliose  bases  make  up  its  surface,  and  whose  com- 
mon altitude  is  the  radius  of  the  sphere. 

801.  Cor.  5.  The  volume  of  the  cylinder  circumscribed 
about  a  sphere  =  27rR\ 

Therefore,  the  volume  of  a  sphei^e  is  equal  to  tico-thirds 
the  volume  of  the  circumscribing  cylinder. 

Note.— It  was  Archimedes  who  discovered  that  the  surface  and  volume  of  the 
sphere  are  each  two-thirds  of  that  of  the  circumscribing  cylinder. 

EXERCISES. 

L  Show  that  the  volumes  of  spherical  sectors  of  the  same 
sphere,  or  equal  spheres,  are  to  each  other  as  the  altitudes 
of  the  zones  which  form  their  bases. 

3.  Find  the  volume  of  a  spherical  sector,  if  the  altitude 
of  the  zone  which  forms  its  base  is  3  feet,  and  the  radius  of 
the  sphere  is  G  feet. 


384 


SOLID   GEOMETUY. 


Proposition   12.    Tiieorem. 

802.  Tlie  volume  of  a  splierical  segiuent  is  equal  to  half 
the  sum  of  its  bases  m,ultipUed  hy  its  altitude,  j^liis  the 
volume  of  a  sphere  of  which  this  altitude  is  the  diameter. 

Hyp.  Let  AB  be  the  arc  of  a  O,  and 
CD  the  projection  of  the  chord  AB  on 
the  diameter  OM. 

Let  AC  =  r',  BD  =  r,  CD  =//,  OA=R, 
and  denote  the  volume  of  the  segment 
generated  by  revolving  the  circular  seg- 
ment AHBDC  about  OM  by  V. 

To  prove  V  ^  \7th{r^  +  r")  +  ^7th\ 

Proof.  Draw  the  radii  OA,  OB. 

The  volume  generated  by  AHBDC  is  the  sum  of  the 
spherical  sector  generated  by  OAHB  and  the  cone  gener- 
ated by  OAC,  diminished  by  the  cone  generated  by  OBD. 

Vol.  sph.  sect.  OAHB  =  %7tWh.  (800) 


Vol.   of  cone      OAC  =  jTrr^OC. 


(778) 


Vol.  of  cone      OBD  =  i;rr''OD. 

.-.  V  =  fTTRVi  +  \7tr''00  -  \7rr''  .  OD. 
But  7^  =  OC  -  OD,  R'  -  r"  =  OC',  and  R''  -  7-'  =  OD 

.-.  V  =  TtWh  -  j7r(0C'  -  OD') 

=  iTthl^U'  -  (OC'  +  OC .  OD  +  OD')] 
Since  OC  -  OD  =  h, 

...  00'  -  20C  .  OD  +  OD'  =  h%         and 

.• .  OC''  +    OC  .  OD  +  OD'  =  f (OC'  +  OD')  -  |-' 


V=|;r7*[i(r'+r")+|'] 

=  i7rh(y'  +  r")  +  Jrf. 

Q.E.D. 


BOOK  IX.— THE  SPHERICAL  SEGMENT.         385 

803.  Cor.  1.  If  the  segment  has  but  one  base,  as  the 
volume  generated  by  MABD,  the  radius  r'  =  0,  and  we 
have 

Therefore,  the  volume  of  a  spherical  segment  of  one  base 
is  equal  to  half  the  cylinder  having  the  same  base  and  the 
same  altitude,  plus  the  sphere  of  which  this  altitude  is  the 
diameter, 

804.  Cor.  2.  When  the  segment  has  but  one  base, 

r'  =  (2R  -  h)h, 
.  • .  V  =  l7rh\2^  -  h)  +  ^7t¥  (803) 

=  7rh\^  -  ^h), 

805.  Cor.  3.  Let  V  denote  the  volume  of  a  frustum  of 
a  cone  generated  by  the  trapezoid  ABDO  about  OM,  and  v 
the  volume  generated  by  the  circular  segment  AHB. 

Then  V  =  \7rh(r''  +  r''  +  rr'),  (782) 

and  V   =i7rh(r'  +  r'')  +  }7rh\  (802) 

Subtracting,  v  =  ^7r7i(r'  +  r"  +  h'  —  2rr') 

=  i7rAB\  h. 

Therefore,  the  volume  generated  by  a  circular  segment 
revolving  about  a  diameter  exterior  to  its  surface,  is  equal  to 
one^sixth  of  the  cylinder  whose  radius  is  the  chord  of  the 
scg?nent  a7id  lohose  altitude  is.  the  projection  of  this  chord 
on  the  axis. 


886  SOLID  GEOMETRY, 


exercises. 
Theorems. 


1.  The  lateral  area  of  a  cylinder  of  revolution  is  equal  to 
the  area  of  a  circle  whose  radius  is  a  mean  proportional  be- 
tween the  altitude  of  the  cylinder  and  the  diameter  of  its 


2.  If  the  slant  height  of  a  cone  of  revolution  is  equal  to 
the  diameter  of  its  base,  its  lateral  area  is  double  the  area 
of  its  base. 

3.  The  volume  of  a  cylinder  of  revolution  is  equal  to  the 
product  of  its  lateral  area  by  half  its  radius. 

4.  A  plane  through  two  elements  of  a  cylinder  of  revo- 
lution cuts  the  base  in  a  chord  which  subtends  at  its  centre 

7t 

an  angle  of  -  :  compare  the  lateral  areas  of  the  two  parts 
o 

of  the  cylinder. 

5.  A  rectangle  revolves  successively  about  two  adjacent 
sides  whose  lengths  are  a  and  h  :  compare  the  volumes  of 
the  two  cylinders  that  are  generated. 

6.  The  two  legs  of  a  right  triangle  are  a  and  h  :  find  the 
area  of  the  surface  generated  when  the  triangle  revolves 
about  its  hypotenuse. 

7.  Prove  that  a  sphere  may  be  inscribed  in  a  cylinder  of 
revolution,  and  that  it  will  touch  it  along  the  circumfer- 
ence of  a  great  circle. 

8.  The  lateral  area  of  a  given  cone  of  revolution  is  double 
the  area  of  its  base  :  find  the  ratio  of  its  altitude  to  the 
radius  of  its  base. 

9.  On  each  base  of  a  frustum  of  a  cone  of  revolution,  a 
cone  stands  having  its  vertex  in  the  centre  of  the  other 
base  :  find  the  radius  of  the  circle  of  intersection  of  the 
two  cones,  the  radii  of  the  bases  being^^  and  r^. 


BOOK  IX.-EXEIWISES.     TIIEOUEMS.  387 

10.  If  the  altitude  of  a  cylinder  of  revolution  be  equal  to 
the  diameter  of  its  base,*  the  volume  is  equal  to  the  product 
of  its  total  area  by  one-third  of  its  radius. 

11.  If  the  slant  height  of  a  cone  of  revolution  be  equal 
to  the  diameter  of  its  base,t  its  total  area  is  to  the  area 
of  the  inscribed  sphere  as  9  :  4. 

12.  In  a  frustum  of  a  cone  of  revolution  the  inclination 
of  the  slant  height  to  one  base  is  45°:  find  the  lateral  area, 
the  radii  of  the  bases  being  r,  and  i\ . 

13.  If  the  radius  of  a  sphere  is  bisected  at  right  angles 
by  a  plane,  the  two  zones  into  which  the  surface  of  the 
sphere  is  divided  are  to  each  other  as  3  :  1. 

14.  If  a  cylinder  and  cone,  each  equilateral,  be  inscribed 
in  a  sphere,  the  total  area  of  the  cylinder  is  a  mean  pro- 
portional between  the  total  area  of  the  cone  and  the  area 
of  the  sphere.  The  same  is  true  of  the  volumes  of  these 
bodies. 

15.  If  a  cylinder  and  cone,  each  equilateral,  be  circum- 
scribed about  a  sphere,  the  total  area  of  the  cylinder  is  a 
mean  proportional  between  the  total  area  of  the  cone  and 
the  area  of  the  sphere.     The  same  is  true  of  the  volumes. 

16.  A  cone  of  revolution  whose  vertical  angle  is  60°,  is 
circumscribed  about  a  sphere:  compare  the  area  of  the 
sphere  and  the  lateral  area  of  the  cone.  Compare  their 
volumes. 

17.  The  base  of  a  cone  is  equal  to  a  great  circle  of  a 
sphere,  and  the  altitude  of  the  cone  is  equal  to  a  diameter 
of  the  sphere :  compare  the  volumes  of  the  cone  and  sphere. 

18.  The^volume  of  a  cone  of  revolution  is  equal  to  the 
area  of  its  generating  rectangle  multiplied  by  the  circum- 
ference generated  by  the  point  of  intersection  of  the  diago- 
nals of  the  rectangle. 

19.  The  Volume  of  a  sphere  is  to  the  volume  of  the  cir- 
cumscribed cube  as  ;r  :  6. 

*  Called  an  equilateral  cylinder. 
t  An  equilateral  cone. 


388  SOLID  GEOMETRY. 

20.  The  volume  of  a  sphere  is  to  the  volume  of  the  in- 
scribed cube  as  7r  :  2. 

21.  If  d  is  the  distance  of  a  point  P  from  the  centre  of 
a  sphere  whose  radius  is  E,  the  sum  of  the  squares  of  the 
six  segments  of  three  chords  at  right  angles  to  each  other 
passing  through  P  is  6R*  —  2iV, 

22.  If  h  is  the  height  of  an  aeronaut,  and  R  is  the  radius 

of  the  earth,  the  extent  of  surface  visible  =  ^D~r~7~  • 

\i  -\-  fi 

Numerical  Exercises. 

23.  Find  the  lateral  area,  total  area,  and  volume,  of  a 
cylinder  of  revolution,  the  radius  of  the  base  being  4  and 
the  altitude  10. 

24.  Find  the  lateral  area,  total  area,  and  volume,  of  a 
cone  of  revolution,  the  radius  of  the  base  being  4  and  the 
altitude  10. 

25.  Find  the  lateral  area,  total  area,  and  volume,  of  a 
frustum  of  a  cone  of  revolution,  the  radii  of  the  bases  being 
7  and  2  and  the  altitude  3. 

26.  Find  the  lateral  area  of  a  frustum  of  a  cone  of  revo- 
lution, the  radii  of  the  bases  being  21  and  6  inches  and  the 
altitude  36  inches.  Ans.  3308.1  cu.  ins. 

27.  Find  the  volume  of  a  frustum  of  a  cone  of  revolu- 
tion, the  radii  of  the  bases  being  4  and  2  feet  and  the  alti- 
tude 9  feet.  Ans,  65.97  cu.  ft. 

28.  The  slant  height  of  a  cone  of  revolution  is  4  feet  : 
how  far  from  the  vertex  must  the  slant  height  be  cut  by  a 
plane  parallel  to  the  base  that  the  lateral  area  may  be 
divided  into  two  equivalent  parts? 

29.  The  altitude  of  a  cone  of  revolution  is  equal  to  the 
diameter  of  its  base :  find  the  ratio  of  the  area^  of  the  base 
to  the  lateral  area. 

30.  Find  the  volume  of  an  equilateral  cylinder  in  terms 
of  its  total  area, 

See  Ex.  (10). 


BOOK  IX.—NVMEIUCAL  EXERCISES.  889 

31.  The  lateral  area  of  a  cylinder  of  revolution  is  116f 
sq.  ft.,  and  the  altitude  is  14  feet:  find  the  diameter  of  its 
base.  Ans.  2.65  ft. 

32.  The  volume  of  a  cylinder  is  15.7  cu.  ft.,  and  the 
diameter  of  its  base  is  2  feet:  find  its  altitude.  A7is,  5  feet. 

33.  The  lateral  area  of  a  cylinder  of  revolution  is  TOtt 
and  its  volume  is  175;r:  find  its  height  and  the  radius  of 
its  base. 

34.  The  lateral  area  of  a  cone  of  revolution  is  60 ;r  and 
its  slant  height  is  12 :  find  its  volume. 

35.  Find  the  total  area  of  a  frustum  of  a  cone  of  revolu- 
tion, the  radii  of  its  bases  being  9  and  4  feet  and  its  height 
12  feet.  Ans.  ^Q^Qn. 

36.  The  volume  of  a  frustum  of  a  cone  of  revolution  is 
920  cu.  ft.,  and  its  height  is  12  feet :  find  the  radii  of  its 
bases  if  their  sum  is  8  feet. 

37.  Find  the  number  of  cubic  feet  in  a  log  12  feet  long 
and  6f  feet  in  diameter.  Ans,  418.88  cu.  ft. 

38.  Find  the  number  of  cubic  feet  in  the  trunk  of  a  tree, 
70  feet  long,  the  diameters  of  its  ends  being  10  and  7  feet. 

39.  How  many  square  inches  of  sheet-iron  does  it  take 
to  make  a  joint  of  6-inch  stovepipe  2^  feet  long,  allowing 
an  inch  and  a  half  for  the  seam  ? 

•  40.  The  heights  of  two  cylinders  of  revolution  of  equal 
volumes  are  as  9  :  16;  the  diameter  of  one  of  them  is  6 
feet:  find  the  diameter  of  the  other. 

41.  A  cylinder  of  revolution  whose  base  is  5  feet  in  di- 
ameter and  a  cone  of  revolution  whose  base  is  6  feet  in  di- 
ameter, have  equal  volumes:  the  height  of  the  cone  is  10 
feet,  find  the  height  of  the  cylinder. 

42.  The  height  of  a  frustum  of  a  cone  of  revolution  is 
6  feet,  and  the  diameters  of  its  bases  are  3  and  2  feet :  find 
the  height  of  a  cylinder  of  revolution  of  the  same  volume 
as  the  frustum,  and  whose  base  is  equal  to  the  mid-section 
of  the  frustum. 

See  (777). 


390  SOLIB   GEOMETRY. 

43.  The  volumes  of  two  equilateral  cylinders  are  to  each 
other  as  3  :  4  :  find  the  ratio  of  their  heights. 

44.  The  volumes  of  two  similar  cones  are  27  cubic  feet 
and  216  cubic  feet,  and  the  height  of  the  first  is  9  feet: 
find  the  height  of  the  other. 

45.  The  volumes  of  two  similar  cones  of  revolution  are 
to  each  other  as  512  :  729  :  find  the  ratio  of  their  lateral 
areas. 

46.  The  slant  heights  of  two  similar  cones  of  revolution 
are  to  each  other  as  3:5:  find  the  ratio  of  their  lateral 
areas,  and  of  their  volumes. 

47.  The  height  of  a  frustum  of  a  cone  is  f  the  height  of 
the  complete  cone :  find  the  ratio  of  the  volume  of  the  frus- 
tum to  that  of  the  cone. 

48.  The  total  areas  of  two  similar  cylinders  of  revolu- 
tion are  to  each  other  as  25  :  49  :  find  the  ratio  of  their 
volumes. 

49.  The  altitude  of  a  cone  of  revolution  is  10,  and  its 
slant  height  is  14:  find  the  total  area  of  the  inscribed 
cylinder  whose  altitude  is  6. 

50.  The  volume  of  a  cone  of  revolution  is  392,  and  its 
slant  height  is  to  the  diameter  of  its  base  as  100  :  56:  find 
its  altitude  and  the  diameter  of  its  base. 

51.  Find  the  surface  and  volume  of  a  sphere  whose  di- 
ameter is  (1)  16  inches;  and  (2)  17  inches. 

Ans.  (1)  804};  2144f :  (2)  908;  2572.45. 

52.  Find  the  diameter  of  a  sphere  if  the  surface  is  (1) 
1809  square  inches;  (2)  616  square  inches;  and  (3)  9856 
square  inches. 

53.  The  volume  of  a  sphere  is  113:  find  its  diameter  and 
its  surface. 

54.  The  volume  of  a  sphere  is  776 ;r:  find  its  diameter 
and  its  surface. 

55.  The  surface  of  a  sphere  is  7847r:  find  its  radius  and 
its  volume. 


BOOK  IX.— NUMERICAL  EXERCISES.  391 

56.  The  volume  of  a  sphere  is  2G8.08  cubic  inches:  find 
the  altitude  of  the  circumscribing  cylinder. 

Ans,  8  inches. 

57.  The  surface  of  a  given  sphere  has  the  same  numer- 
ical value  as  the  circumference  of  a  great  circle:  find  the 
numerical  value  of  its  radms. 

58.  Find  the  surface  of  a  zone  of  one  base,  its  altitude 
being  10  feet,  and  the  diameter  of  the  sphere  100  feet. 

59.  Find  the  surface  of  a  zone,  its  altitude  being  6  feet, 
and  the  diameter  of  the  sphere  20  feet. 

60.  Find  the  height  of  a  zone  whose  area  is  equal  to  that 
of  a  great  circle  in  a  sphere  of  radius  r. 

61.  Assuming  the  earth  to  be  a  sphere  with  a  radius  of 
4000  miles,  and  the  altitudes  of  the  torrid  and  the  temper- 
ate zones  to  be  3200  and  2052  miles  respectively,  find  the 
areas  of  these  two  zones. 

62.  The  surface  and  volume  of  a  given  sphere  are  ex- 
pressed by  the  same  number:  find  its  diameter. 

63.  Find  the  weight  of  an  iron  shell  4  inches  in  diameter, 
the  iron  being  1  inch  thick,  and  weighing  ^  of  a  pound  to 
the  cubic  inch.  Aiis.  7i  lbs. 

64.  A  sphere  of  radius  r  is  cut  by  a  plane  so  that  the 
area  of  the  greater  zone  is  a  mean  proportional  between 
the  area  of  the  smaller  zone  and  the  area  of  the  sphere. 

65.  Find  the  surface  of  a  sphere  inscribed  in  a  cube 
whose  surface  is  216. 

66.  Find  the  volume  of  a  sphere  circumscribed  about  a 
cube  whose  volume  is  64, 

67.  If  an  iron  ball  8  inches  in  diameter  weighs  72  pounds, 
find  the  weight  of  an  iron  shell  10  inches  in  diameter,  the 
iron  being  2  inches  thick. 

68.  A  cone  of  revolution,  the  radius  of  whose  base  is  12, 
is  inscribed  in  a  sphere  of  radius  20:  find  the  volume  of  the 
cone. 

69.  How  high  above  the  earth  must  a  person  be  raised 
in  order  that  he  may  see  one-fifth  of  its  surface? 


392  SOLID  GEOMETRY. 

70.  How  much  of  tlie  earth's  surface  would  a  man  see  if 
he  were  raised  to  the  height  of  the  diameter  above  it? 

71.  Find  the  vohime  of  a  spherical  sector,  if  the  altitude 
of  the  zone  which  forms  its  base  is  3  feet,  and  the  radius  of 
the  sphere  is  5  feet. 

72.  Find  the  volume  of  a  spherical  sector,  if  the  area  of 
the  zone  which  forms  its  base  is  5  square  feet,  and  the 
radius  of  the  sphere  is  2  feet. 

73.  Find  the  volume  of  the  spherical  sector  generated  by 
revolving  a  circular  sector  about  an  axis  in  its  plane  per- 
pendicular to  one  of  its  limiting  radii,  the  radius  of  the 
circuhir  sector  being  6  and  its  central  angle  30°. 

74.  Find  the  volume  of  a  triangular  spherical  pyramid,  if 
the  angles  of  the  spherical  triangle  which  forms  its  base  are 
each  120°,  and  the  radius  of  the  spliere  is  10  feet. 

75.  Find  the  volume  of  a  quadrangular  spherical  pyramid, 
if  the  planes  of  the  four  faces  of  the  pyramid  make  with 
each  other  angles  of  80°,  100°,  120°,  150°,  and  a  lateral 
edge  of  the  pyramid  is  4  feet. 

76.  Find  the  volume  of  a  spherical  segment,  the  radii  of 
whose  bases  are  3  and  5,  and  whose  altitude  is  4. 

77.  Find  the  volume  of  a  spherical  segment,  the  diam- 
eters of  whose  bases  are  24  and  20  inches,  and  whose  alti- 
tude is  4  inches. 

78.  Find  the  volume  of  a  spherical  segment  of  one  base 
whose  altitude  is  4  feet,  the  radius  of  the  sphere  being  10 
feet. 

79.  Find  the  volume  of  a  spherical  segment  of  one  base 
whose  altitude  is  6  inches  and  the  diameter  of  its  base  16 
inches. 

80.  A  sphere,  2  feet  in  diameter,  is  cut  by  two  parallel 
planes,  one  at  3  and  the  other  at  9  inches  from  the  centre : 
find  the  volume  of  the  segment  included  between  them. 

81.  Find  the  volume  of  a  spherical  segment  of  one  base 
whose  altitude  is  20  feet  and  the  diameter  of  its  base  60 
feet. 


BOOK  IX.-NUMERICAL  EXERCISES.  393 

82.  The  radius  of  a  sphere  is  6  inches:  find  the  area  of  a 
spherical  triangle  whose  angles  are  95°,  100°,  120°. 

83.  The  radius  of  a  sphere  is  7  feet :  find  (1)  the  area  of 
a  lune  whose  angle  is  30°,  and  (2)  the  volume  of  a  wedge 
whose  angle  is  3G°. 

84.  A  sphere  6  inches  in  diameter  has  a  hole  bored 
through  its  centre  with  a  3-inch  auger  :  find  the  remain- 
ing volume. 

85.  A  boiler  is  a  cylinder  with  hemispherical  ends ;  its 
total  length  is  20  feet  and  circumference  11  feet:  find  its 
surface  and  the  quantity  of  water  required  to  fill  it  half  full. 

86.  A  cone  is  circumscribed  about  a  sphere,  and  its 
height  is  double  the  diameter  of  the  sphere :  show  that  the 
total  surface  and  the  volume  of  the  cone  are  respectively 
double  those  of  the  sphere. 

87.  A  circle,  radius  r,  revolves  about  a  line  in  its  plane 
and  at  a  distance  d  from  its  centre  :  find  the  volume  of  the 
ring  which  the  circle  generates. 


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